A+B Problem II

A+B Problem II

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

21 2112233445566778899 998877665544332211

样例输出

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

大数乘法问题，学习了。

#include<stdio.h>#include<string.h>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--)    {              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]);           if(num[k]/10)           {               num[k-1]++;               num[k]%=10;           }       }       while(al>0)       {            num[k--]+=shu(a[--al]);            if(num[k+1]/10)           {               num[k]++;               num[k+1]%=10;           }       }       while(bl>0)       {            num[k--]+=shu(b[--bl]);            if(num[k+1]/10)           {               num[k]++;               num[k+1]%=10;           }       }       printf("Case %d:\n",j++);       printf("%s + %s = ",a,b);       for(i=0;i<=t;i++)       {           if(i==0&&num[i]==0)           i++;           printf("%d",num[i]);       }       printf("\n");       if(n!=0) {       printf("\n");       }   }   return 0;}