大数a+b

E - 大数加法

HDU - 1002        

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

题意：大数a+b；

#include<stdio.h>#include<string.h>char a[1000+13],b[1000+13];int c[1000+10],d[1000+13],e[1000+13],f[1000+13];int max(int a,int b){if(a>b) return a;return b;}int main(){int i,t,m,n,x,len1,len2,len3,j=0;scanf("%d\n",&t);int k=t;while(t--){if(k>t+1){printf("\n");}x=0;memset(e,0,sizeof(e));memset(c,0,sizeof(c));memset(d,0,sizeof(d));scanf("%s %s",a,b);len1=strlen(a);len2=strlen(b);m=0;n=0;for(i=len1-1;i>=0;i--){c[n++]=a[i]-'0';    }for(i=len2-1;i>=0;i--){d[m++]=b[i]-'0';}for(i = 0; i < max(len1, len2) + 1; i++)    {            e[i] = 0;        }for(i=0;i<max(len1,len2)+1;i++){e[i] = c[i] + d[i] + e[i];            if(e[i] >= 10){                e[i]-= 10;                e[i + 1]++;            }}for(i=max(len1,len2)+1;i>=0;i--){x++;if(e[i]!=0){break;}}j++;printf("Case %d:\n",j);printf("%s + %s = ",a,b);for(i=max(len1,len2)-x+2;i>=0;i--){printf("%d",e[i]);}printf("\n");}return 0;}