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部分ACM题目与答案

1001 Sum Problem (2)

1089 A+B for Input-Output Practice (I) (3)

1090 A+B for Input-Output Practice (II) (3)

1091 A+B for Input-Output Practice (III) (3)

1092 A+B for Input-Output Practice (IV) (3)

1093 A+B for Input-Output Practice (V) (3)

1094 A+B for Input-Output Practice (VI) (3)

1095 A+B for Input-Output Practice (VII) (3)

1096 A+B for Input-Output Practice (VIII) (3)

2000 ASCII码排序 (3)

2001计算两点间的距离 (3)

2002计算球体积 (3)

2003求绝对值 (3)

2004成绩转换 (3)

2005第几天？ (3)

2006求奇数的乘积 (3)

2007平方和与立方和 (3)

2008数值统计 (3)

2009求数列的和 (3)

2010水仙花数 (3)

2011多项式求和 (3)

2012素数判定 (3)

2014青年歌手大奖赛_评委会打分 (3)

2015偶数求和 (3)

2016数据的交换输出 (3)

2017字符串统计 (3)

2019数列有序! (3)

2020绝对值排序 (3)

2021发工资咯：） (3)

2033人见人爱A+B (3)

2039三角形 (3)

2040亲和数 (3)

1001 Sum Problem

Problem Description

Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.

Input

The input will consist of a series of integers n, one integer per line.

Output

For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

Sample Input

100

Sample Output

5050

Author

DOOM III

解答：

#include

main()

{

int n,i,sum;

sum=0;

while((scanf("%d",&n)!=-1))

{

sum=0;

for(i=0;i<=n;i++)

sum+=i;

printf("%d\n\n",sum);

}

1089 A+B for Input-Output Practice

(I)

Problem Description

Your task is to Calculate a + b.

Too easy?! Of course! I specially designed the problem for acm beginners.

You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.

Input

The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

Output

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

Sample Input

1 5

10 20

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

main()

{

int a,b;

while(scanf("%d%d",&a,&b)!=EOF) printf("%d\n",a+b);

}

1090 A+B for Input-Output Practice

(II)

Problem Description

Your task is to Calculate a + b.

Input

Input contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.

Output

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

Sample Input

1 5

10 20

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

#define M 1000

void main()

{

int a,b,n,j[M],i;

//printf("please input n:\n");

scanf("%d",&n);

for(i=0;i

{

scanf("%d%d",&a,&b);

//printf("%d %d",a,b);

j[i]=a+b;

}

i=0;

while(i

{

printf("%d",j[i]);

i++;

printf("\n");

}

1091A+B for Input-Output Practice

(III)

Problem Description

Your task is to Calculate a + b.

Input

Input contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.

Output

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

Sample Input

1 5

10 20

0 0

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

main()

{

int a,b;

scanf("%d %d",&a,&b);

while(!(a==0&&b==0))

{

printf("%d\n",a+b);

scanf("%d %d",&a,&b);

}

1092A+B for Input-Output Practice

(IV)

Problem Description

Your task is to Calculate the sum of some integers.

Input

Input contains multiple test cases. Each test case contains a integer N, and then N integers follow in the same line. A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each group of input integers you should output their sum in one line, and with one line of output for each line in input.

Sample Input

4 1 2 3 4

5 1 2 3 4 5

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

int main()

{

int n,sum,i,t;

while(scanf("%d",&n)!=EOF&&n!=0)

{

sum=0;

for(i=0;i

{

scanf("%d",&t);

sum=sum+t;

}

printf("%d\n",sum);

}

1093 A+B for Input-Output Practice

(V)

Problem Description

Your task is to calculate the sum of some integers.

Input

Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

Output

For each group of input integers you should output their sum in one line, and with one line of output for each line in input.

Sample Input

4 1 2 3 4

5 1 2 3 4 5

Sample Output

Author

lcy

解答：

#include

main()

{

int n,a,b,i,j,sum;

sum=0;

while(scanf("%d\n",&n)!=-1)

{

for(i=0;i

{

scanf("%d",&b);

for(j=0;j

{

scanf("%d",&a);

sum+=a;

}

printf("%d\n",sum);

sum=0;

}

1094 A+B for Input-Output Practice

(VI)

Problem Description

Your task is to calculate the sum of some integers.

Input

Input contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.

Output

For each test case you should output the sum of N integers in one line, and with one line of output for each line in input.

Sample Input

4 1 2 3 4

5 1 2 3 4 5

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

main()

{

int n,a,b,i,j,sum;

sum=0;

while(scanf("%d\n",&n)!=-1)

{

for(j=0;j

{

scanf("%d",&a);

sum+=a;

}

printf("%d\n",sum);

sum=0;

}

[ Copy to Clipboard ][ Save to File]

1095A+B for Input-Output Practice

(VII)

Problem Description

Your task is to Calculate a + b.

Input

The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.

Output

For each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.

Sample Input

1 5

10 20

Sample Output

Author

lcy

Recommend

JGShining

解答：

#include

main()

{

int a,b;

while(scanf("%d%d",&a,&b)!=EOF)

printf("%d\n\n",a+b);

}

1096 A+B for Input-Output Practice

(VIII)

Problem Description

Your task is to calculate the sum of some integers.

Input

Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

Output

For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.

Sample Input

4 1 2 3 4

5 1 2 3 4 5

3 1 2 3

Sample Output

Author

lcy

Recommend

JGShining

解答：

int main()

{

int a,b,i,j,l[1000],k;

scanf("%d",&i);

getchar();

for(j=1;j<=i;j++)

l[j]=0;

for(j=1;j<=i;j++)

{

scanf("%d",&a);

getchar();

for(k=1;k<=a;k++)

{

scanf("%d",&b);

getchar();

l[j]+=b;

}

for(j=1;j<=i-1;j++)

printf("%d\n\n",l[j]);

printf("%d\n",l[i]);

}

2000 ASCII码排序Problem Description

输入三个字符后，按各字符的ASCII码从小到大的顺序输出这三个字符。

Input

输入数据有多组，每组占一行，有三个字符组成，之间无空格。

Output

对于每组输入数据，输出一行，字符中间用一个空格分开。

Sample Input

qwe

asd

zxc

Sample Output

e q w

a d s

c x z

Author

lcy

Source

C语言程序设计练习（一）

Recommend

JGShining

解答：

#include

main()

(完整版)杭电acm部分答案

Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 #include void main() { int a,b; while(scanf("%d %d",&a,&b)!=EOF) { printf("%d\n",a+b); } } Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer. Sample Input 1 100 Sample Output

杭州电子科技大学OJ题目分类

杭州电子科技大学OJ题目分类 1001 整数求和水题 1002 C语言实验题——两个数比较水题 1003 1、2、3、4、5... 简单题 1004 渊子赛马排序+贪心的方法归并 1005 Hero In Maze 广度搜索 1006 Redraiment猜想数论：容斥定理 1007 童年生活二三事递推题 1008 University 简单hash 1009 目标柏林简单模拟题 1010 Rails 模拟题（堆栈） 1011 Box of Bricks 简单题 1012 u Calculate e 简单数学计算 1013 STAMPS 搜索or动态规划 1014 Border 模拟题 1015 Simple Arithmetics 高精度计算 1016 Shoot-out 博弈+状态压缩DP 1017 Tour Guide 1018 Card Trick 简单题 1019 Necklace Decomposition 贪心 1020 Crashing Robots 模拟题 1021 Electrical Outlets 简单题 1022 Watchdog 简单题 1023 Taxi Cab Scheme 图论：最小路径覆盖--->最大二分匹配1024 Pseudo-random Numbers 数论 1025 Card Game Cheater 简单题 1026 Investment 动态规划 1027 Pipes 1028 SETI 数学:高斯消元法 1029 Minimax Triangulation 计算几何 1030 Unequalled Consumption 母函数 1031 Declaration of Content 1032 Laserbox 搜索：DFS 1033 Bowlstack 1034 Pesky Heroes 1035 Reduced ID Numbers 暴力 1036 Tantrix 1037 Guardian of Decency 图论：匈牙利算法求二分图的最大匹配1038 Up the Stairs 简单数学题 1039 Sudoku 搜索：DFS 1040 The SetStack Computer 1041 Pie 二分法 1042 Ticket to Ride 动态规划 1043 The Bookcase 动态规划

杭电ACM水题题目及代码

1001 #include int main() { int i,a,j;double sum; while(scanf("%d",&a)!=EOF) { sum=0; for(j=1;j<=a;j++) { sum+=j; } printf("%.0lf\n\n",sum); } return 0; } 1002 A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 69615 Accepted Submission(s): 12678 Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases. Sample Input 2 1 2 112233445566778899 998877665544332211 Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 Author Ignatius.L #include #include int main(){ char str1[1001], str2[1001]; int t, i, len_str1, len_str2, len_max, num = 1, k; scanf("%d", &t); getchar(); while(t--){ int a[1001] = {0}, b[1001] = {0}, c[1001] = {0}; scanf("%s", str1); len_str1 = strlen(str1);

杭电OJ题目分类

杭州电子科技大学OJ题目分类The Soul with Bone .: 1001 整数求和水题 1002 C语言实验题——两个数比较水题 1003 1、2、3、4、5... 简单题 1004 渊子赛马排序+贪心的方法归并 1005 Hero In Maze 广度搜索 1006 Redraiment猜想数论：容斥定理 1007 童年生活二三事递推题 1008 University 简单hash 1009 目标柏林简单模拟题 1010 Rails 模拟题（堆栈） 1011 Box of Bricks 简单题 1012 IMMEDIATE DECODABILITY Huffman编码 1013 STAMPS 搜索or动态规划 1014 Border 模拟题 1015 Simple Arithmetics 高精度计算 1016 Shoot-out 博弈+状态压缩DP 1017 Tour Guide 1018 Card Trick 简单题 1019 Necklace Decomposition 贪心

1020 Crashing Robots 模拟题 1021 Electrical Outlets 简单题 1022 Watchdog 简单题 1023 Taxi Cab Scheme 图论：最小路径覆盖--->最大二分匹配1024 Pseudo-random Numbers 数论 1025 Card Game Cheater 简单题 1026 Investment 动态规划 1027 Pipes 1028 SETI 数学:高斯消元法 1029 Minimax Triangulation 计算几何 1030 Unequalled Consumption 母函数 1031 Declaration of Content 1032 Laserbox 搜索：DFS 1033 Bowlstack 1034 Pesky Heroes 1035 Reduced ID Numbers 暴力 1036 Tantrix 1037 Guardian of Decency 图论：匈牙利算法求二分图的最大匹配1038 Up the Stairs 简单数学题 1039 Sudoku 搜索：DFS 1040 The SetStack Computer 1041 Pie 二分法

杭电acm部分题目及答案答案

自己刷的题这是我在杭电做题的记录，希望我的分享对你有帮助！！！ 1001 Sum Problem***********************************************************1 1089 A+B for Input-Output Practice (I)********************************2 1090 A+B for Input-Output Practice (II)********************************5 1091A+B for Input-Output Practice (III)****************************************7 1092A+B for Input-Output Practice (IV)********************************8 1093 A+B for Input-Output Practice (V)********************************10 1094 A+B for Input-Output Practice (VI)***************************************12 1095A+B for Input-Output Practice (VII)*******************************13 1096 A+B for Input-Output Practice (VIII)******************************15 How to Type***************************************************************16 1001 Sum Problem Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

整理出ACM所有题目及答案

1111111杭电： 1000 A + B Problem (4) 1001 Sum Problem (5) 1002 A + B Problem II (6) 1005 Number Sequence (8) 1008 Elevator (9) 1009 FatMouse' Trade (11) 1021 Fibonacci Again (13) 1089 A+B for Input-Output Practice (I) (14) 1090 A+B for Input-Output Practice (II) (15) 1091 A+B for Input-Output Practice (III) (16) 1092 A+B for Input-Output Practice (IV) (17) 1093 A+B for Input-Output Practice (V) (18) 1094 A+B for Input-Output Practice (VI) (20) 1095 A+B for Input-Output Practice (VII) (21) 1096 A+B for Input-Output Practice (VIII) (22) 1176 免费馅饼 (23) 1204 糖果大战 (25) 1213 How Many Tables (26) 2000 ASCII码排序 (32) 2001 计算两点间的距离 (34) 2002 计算球体积 (35) 2003 求绝对值 (36) 2004 成绩转换 (37) 2005 第几天？ (38) 2006 求奇数的乘积 (40) 2007 平方和与立方和 (41) 2008 数值统计 (42) 2009 求数列的和 (43) 2010 水仙花数 (44) 2011 多项式求和 (46) 2012 素数判定 (47) 2014 青年歌手大奖赛_评委会打分 (49) 2015 偶数求和 (50) 2016 数据的交换输出 (52) 2017 字符串统计 (54) 2019 数列有序! (55) 2020 绝对值排序 (56) 2021 发工资咯：） (58) 2033 人见人爱A+B (59) 2037 今年暑假不AC (61) 2039 三角形 (63) 2040 亲和数 (64)

杭电ACM试题详细分类,杭电oj详细分类,hdu详细分类,详细,ACM.doc

杭电ACM试题分类枚举 1002 10041013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 10471048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 11061107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 11971200 1201 1202 1205 1209 1212（大数取模）1216 （链表）1218 1219 1225 1228 12291230 1234 1235 1236 1237 1239 1250 1256 1259 1262 1263 1265 1266 1276 1279 1282 1283 1287 1296 1302 1303 1304 1305 1306 1309 1311 1314 搜索，递归求解 1010 1016 1026 1043（双广）1044 （BFS+DFS） 1045 1067 1072 1104 1175 1180 1195 1208 1226 1238 1240 1241 1242 1258 1271 1312 1317 动态规划 1003 1024 1025 1028 1051 1058 1059 1069 1074 1078 1080 1081 1085 1087 1114 1158 1159 1160 1171 1176 1181 1203 1224 1227 1231 1244 1248 1253 1254 1283 1300 数学，递推，规律 1005 1006 1012 1014 1018 1019 1021 1023 1027 1030 1032 1038 1041 1046 1059 1060 1061 1065 1066 1071（微积分）1097 1098 1099 1100 1108 1110 1112 1124 1130 1131 1132 1134 1141 1143 1152 1155（物理题）1163 1165 1178 1194 1196（lowbit） 1210 1214 1200 1221 1223 1249 1261 1267 1273 1290 1291 1292 1294 1297 1313 1316 数论 1164 1211 1215 1222 1286 1299 计算几何 1086 1115 1147

ACM入门十题(杭电oj)

ACM入门（杭电oj） Hdu 1000 #include #include int main() { int a,b; while(scanf("%d%d",&a,&b)!=EOF) { printf("%d\n",a+b); } } Hdu 1001 #include #include int main() { int n; while(scanf("%d",&n)!=EOF) { printf("%I64d\n\n",(__int64)(1+n)*n/2); } } Hdu 1002 #include #include #include char str1[1005],str2[10005]; int main() { int ca,count=0; scanf("%d",&ca); while(ca--) { scanf("%s%s",str1,str2); int a[1005],i,j; memset(a,0,sizeof(a)); for(i=strlen(str1)-1,j=0;i>=0;i--,j++) a[j]=str1[i]-'0'; for(i=strlen(str2)-1,j=0;i>=0;i--,j++) {

a[j]=a[j]+str2[i]-'0'; a[j+1]=a[j+1]+a[j]/10; a[j]=a[j]%10; } count++; printf("Case %d:\n",count); printf("%s + %s = ",str1,str2); int flag=0; for(i=1004;i>=0;i--) if(flag||a[i]) { printf("%d",a[i]); flag=1; } printf("\n"); if(ca!=0) printf("\n"); } } Hdu 1003 #include #include int a[100005],sum[100005]; int main() { int ca,count=0; scanf("%d",&ca); while(ca--) { int n,i; scanf("%d",&n); for(i=1;i<=n;i++) scanf("%d",&a[i]); sum[1]=a[1]; int r=1,max=a[1]; for(i=2;i<=n;i++) { if(sum[i-1]>0) { sum[i]=sum[i-1]+a[i]; if(sum[i]>max) { max=sum[i]; r=i;

杭电题目acm答案

1001 Sum Problem (2) 1089 A+B for Input-Output Practice (I) (4) 1090 A+B for Input-Output Practice (II) (6) 1091 A+B for Input-Output Practice (III) (8) 1092 A+B for Input-Output Practice (IV) (9) 1093 A+B for Input-Output Practice (V) (11) 1094 A+B for Input-Output Practice (VI) (12) 1095 A+B for Input-Output Practice (VII) (13) 1096 A+B for Input-Output Practice (VIII) (14) 2000 ASCII码排序 (16) 2001计算两点间的距离 (17) 2002计算球体积 (19) 2003求绝对值 (20) 2004成绩转换 (21) 2005第几天 (22) 2006求奇数的乘积 (24) 2007平方和与立方和 (26) 2008数值统计 (27) 2009求数列的和 (28) 2010水仙花数 (29) 2011多项式求和 (31) 2012素数判定 (33) 2014青年歌手大奖赛_评委会打分 (34) 2015偶数求和 (36) 2016数据的交换输出 (38) 2017字符串统计 (40) 2019数列有序! (41) 2020绝对值排序 (43) 2021发工资咯：） (45) 2033人见人爱A+B (46) 2039三角形 (48) 2040亲和数 (49) 姓名：郑春杰班级：电商1001 学号：34

完整word版杭电ACM试题答案

【杭电ACM1000】 A + B Problem Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 # include int main() { int a, b; while(scanf(%d%d, &a, &b)!=EOF) printf(%d\n, a+b); return 0; } 【杭电ACM1001】 Sum Problem Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.

Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer. Sample Input 1 100 Sample Output 1 5050 # include int main() { int n, i, sum = 0; while(scanf(%d, &n)!=EOF) { for(i=1; i<=n; ++i) sum = sum + i; printf(%d\n\n, sum); sum = 0; } return 0; } 【杭电ACM1002】 A + B Problem II Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers

ACM部分练习题目答案

ACM部分习题答案： A + B Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 100972 Accepted Submission(s): 33404 Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 # include Int main() {int x,y,s; while(scanf("%d %d",&x,&y)!=EOF) {s=x+y; printf("%d\n",s);} return 0; } Sum Problem Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 85964 Accepted Submission(s): 19422 Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer. Sample Input 1 100 Sample Output 1 5050 # include int main() {int n; long int s;

杭电ACM部分题答案

1000A + B Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 158161 Accepted Submission(s): 50186 Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 Author HDOJ Statistic | Submit | Discuss | Note #include int main() { int a,b; while(scanf("%d %d",&a,&b)!=EOF) printf("%d\n",a+b); }

1002A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 84367 Accepted Submission(s): 15966 Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases. Sample Input 2 1 2 112233445566778899 998877665544332211 Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 Author Ignatius.L Statistic | Submit | Discuss | Note

杭电题目acm答案

选修课考试作业 1001 Sum Problem ............................................................................................ 错误!未定义书签。1089 A+B for Input-Output Practice (I) .......................................................... 错误!未定义书签。1090 A+B for Input-Output Practice (II) ......................................................... 错误!未定义书签。1091 A+B for Input-Output Practice (III) ........................................................ 错误!未定义书签。1092 A+B for Input-Output Practice (IV) ........................................................... 错误!未定义书签。1093 A+B for Input-Output Practice (V) ......................................................... 错误!未定义书签。1094 A+B for Input-Output Practice (VI) ........................................................ 错误!未定义书签。1095 A+B for Input-Output Practice (VII) .......................................................... 错误!未定义书签。1096 A+B for Input-Output Practice (VIII) ...................................................... 错误!未定义书签。' 2000 ASCII码排序 ............................................................................................ 错误!未定义书签。2001计算两点间的距离.................................................................................. 错误!未定义书签。2002计算球体积 ............................................................................................. 错误!未定义书签。2003求绝对值 ................................................................................................. 错误!未定义书签。2004成绩转换 ................................................................................................. 错误!未定义书签。2005第几天 ..................................................................................................... 错误!未定义书签。2006求奇数的乘积 ......................................................................................... 错误!未定义书签。2007平方和与立方和...................................................................................... 错误!未定义书签。2008数值统计 ................................................................................................. 错误!未定义书签。2009求数列的和 ............................................................................................. 错误!未定义书签。~ 2010水仙花数 ................................................................................................. 错误!未定义书签。2011多项式求和 ............................................................................................. 错误!未定义书签。2012素数判定 ................................................................................................. 错误!未定义书签。2014青年歌手大奖赛_评委会打分................................................................ 错误!未定义书签。2015偶数求和 ................................................................................................. 错误!未定义书签。2016数据的交换输出...................................................................................... 错误!未定义书签。2017字符串统计 ............................................................................................. 错误!未定义书签。2019数列有序! ................................................................................................ 错误!未定义书签。2020绝对值排序.............................................................................................. 错误!未定义书签。2021发工资咯：）............................................................................................ 错误!未定义书签。: 2033人见人爱A+B .......................................................................................... 错误!未定义书签。2039三角形 ..................................................................................................... 错误!未定义书签。2040亲和数 ..................................................................................................... 错误!未定义书签。姓名：郑春杰班级：电商1001

杭电ACM博弈题合集

hdu博弈，这些题都不难。属于博弈简单题。 hdu1846巴什博弈，n%(m+1)==0先手必败。 #include #include #include #include #include using namespace std; int main() { int n,a,b; scanf("%d",&n); while(n--) { scanf("%d%d",&a,&b); if(a%(b+1)==0) printf("second\n"); else printf("first\n"); } return 0; } hdu1847 只要留下两类都是2的指数幂，就是必输状态，然后找规律，发现这两类的和为3的倍数。即有下面的结论。 #include #include #include using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { if(n%3==0) printf("Cici\n"); else printf("Kiki\n"); } return 0; } hdu1848 #include #include

#include #include using namespace std; #define N 1005 int f[N]; int sg[N]; void fun() { int i; f[0]=1; f[1]=1; f[2]=2; for(i=3;;i++) { f[i]=f[i-1]+f[i-2]; if(f[i]>1000) break; } } int dfs(int v) { int i; if(sg[v]!=-1) return sg[v]; bool visit[N]={0}; for(i=1;i<16;i++) { if(v>=f[i]) { int temp=dfs(v-f[i]); visit[temp]=1; } } for(i=0;visit[i];i++); return sg[v]=i; } int main() { fun(); int m,n,p; while(scanf("%d%d%d",&m,&n,&p),m||n||p) { memset(sg,-1,sizeof(sg)); int ans;

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