C语言之大数加法

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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

2
1 2
112233445566778899 998877665544332211

样例输出

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

来源

经典题目

#include<stdio.h>
#include<string.h>
int main()
{
    char str1[1001],str2[1001];
    int t,j=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",str1,str2);
        int a[1001]= {0},b[1001]= {0},c[1001]= {0},i;
        int m,n,max1=0;
        m=strlen(str1);
        n=strlen(str2);
        max1=(m>n?m:n);//m,n大小，输出较大的数的长度
        for(i=0; i<max1; i++)
        {
            a[m-i-1]=str1[i]-'0';
            b[n-i-1]=str2[i]-'0';
        }
        for(i=0; i<max1; i++)
            c[i]=a[i]+b[i];
        for(i=0; i<max1; i++)
        {
            c[i+1]+=c[i]/10;
            c[i]=c[i]%10;
        }
        printf("Case %d:\n%s + %s = ",++j,str1,str2);
        //倒序输出和；
        if(c[max1]==0)
            max1--;
        for(i=max1; i>=0; i--)
            printf("%d",c[i]);
        printf("\n");
    }
    return 0;
}