南阳103 A+B Problem II

 

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

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){    int t;    int la,lb;    int Case=1;    scanf("%d",&t);    while(t--)    {        int j,sum;        char a[1100],b[1100];        int an[1100]={0},bn[1100]={0};        int ans[1100]={0};        scanf("%s %s",a,b);        la=strlen(a);        lb=strlen(b);        j=1;        for(int i=la-1;i>=0;i--)            an[j++]=a[i]-'0';        j=1;        for(int i=lb-1;i>=0;i--)            bn[j++]=b[i]-'0';        sum=0;        j=1;        for(int i=1;i<=((la>lb)?la:lb);i++)        {            sum+=an[i]+bn[i];            ans[j++]=sum%10;            sum/=10;        }            ans[j]=sum;        printf("Case %d:\n",Case++);        printf("%s + %s = ",a,b);        for(int i=((sum==0)?(j-1):j);i>=1;i--)        {            printf("%d",ans[i]);        }        printf("\n");                }    return 0;}