103 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

#include<iostream>#include<cstdio>#include<cstring>using namespace std;void rev(char s1[]){    int i,j,l;    char s2[1005];    l=strlen(s1);    for(i=l-1,j=0;i>=0;i--,j++)        s2[j]=s1[i];    s2[j]='\0';    strcpy(s1,s2);    for(i=j;i<=1000;i++)        s1[i]='0';    s1[1001]='\0';}int main(){    int t,i,j,la,lb,p,q;    char a[1005],b[1005],c[1005];    scanf("%d",&t);    for(i=1;i<=t;i++)    {        scanf("%s%s",a,b);        printf("Case %d:\n",i);        printf("%s + %s = ",a,b);        la=strlen(a);        lb=strlen(b);        rev(a); rev(b);        q=0;        for(j=0;j<=1000;j++)        {            p=(a[j]-'0')+(b[j]-'0')+q;            c[j]=p%10+'0';            q=p/10;        }        for(j=1000;j>=0 && c[j]=='0';j--);        for(;j>=0;j--)            printf("%c",c[j]);        printf("\n");    }}