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

来源

经典题目

上传者

张云聪

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;char a[1100],b[1100];int sum[1100];int main(){int T,i,j,lena,lenb,count=1;cin>>T;while(T--){memset(sum,0,sizeof(sum));int start=0;j=1099;       cin>>a>>b;   lena=strlen(a);   lenb=strlen(b);   for(i=lena-1;i>=0;i--)   {   sum[j]+=a[i]-'0';   j--;   }   j=1099;   for(i=lenb-1;i>=0;i--)   {   sum[j]+=b[i]-'0';   j--;   }   for(i=1099;i>=1;i--)   {   if(sum[i]>=10)   {   sum[i-1]+=sum[i]/10;   sum[i]=sum[i]%10;   }   }   printf("Case %d:\n",count++);   printf("%s + %s = ",a,b);   while(!sum[start] && start<1099)   start++;   for(i=start;i<=1099;i++)   cout<<sum[i];     cout<<endl;}return 0;}