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

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

#include

#include

int shu(char a)

{

return (a-'0');

}

int main(){

char a[1000],b[1000];

int num[1001];

int n,i,j=1,al,bl,k,t;

scanf(%d,&n);

while(n--)

{

getchar();

if(j!=1)

printf(\

);

scanf(%s,a);

al=strlen(a);

scanf(%s,b);

bl=strlen(b);

k=(al>bl)?al:bl;

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

num[i]=0;

t=k;

for(k;al>0&&bl>0;k--)

{

num[k]+=shu(a[--al])+shu(b[--bl]);