hd1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
 
题目大意就是大数相加
 
#include<stdio.h>#include<string.h>int main(){int n,i,j,k,t,lena,lenb,a[1001],b[1001],sum[1001];char sa[1001],sb[1001];scanf("%d",&n);for(k=1;k<=n;++k){printf("Case %d:\n",k);memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(sum,0,sizeof(sum));scanf("%s %s",sa,sb);lena=strlen(sa);lenb=strlen(sb);for(i=lena-1,j=0;i>=0;j++,i--)a[j]=sa[i]-'0';for(i=lenb-1,j=0;i>=0;j++,i--)b[j]=sb[i]-'0';for(i=0;i<1001;++i){sum[i]=sum[i]+a[i]+b[i];if(sum[i]>=10){sum[i]-=10;sum[i+1]+=1;}}printf("%s + %s = ",sa,sb);i=1000;while(sum[i]==0&&i>=0)i--;for(;i>=0;i--)printf("%d",sum[i]);printf("\n");if(k!=n)printf("\n");}return 0;}