大数运算之加法

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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

#include <stdio.h>
#include <string.h>

int main()
{
    int num = 1, n;
    scanf("%d", &n);
    while(n--)
    {
        char s1[1024], s2[1024];
        int x[1024]={0}, y[1024]={0}, sum[1024]={0};

        scanf("%s %s", s1, s2);
        int len1 = strlen(s1);
        int len2 = strlen(s2);

        int i, k=0, j=0, p=0;

        for(i=len1-1; i>=0; i--)
            x[k++] = s1[i] - ‘0‘;

        for(i=len2-1; i>=0; i--)
            y[j++] = s2[i] - ‘0‘;

        int max = k>j?k:j;

        for(i=0; i<=max; i++)
        {
            sum[i] = x[i] + y[i] + p;
            if(sum[i] > 9)
            {
                sum[i] -= 10;
                p = 1;
            }
            else p=0;
        }
        printf("Case %d:\n", num++);
        printf("%s + %s = ", s1, s2);
        if(sum[max]) printf("%d", sum[max]);
        for(i=max-1; i>=0; i--)
            printf("%d", sum[i]);
        printf("\n");
    }

    return 0;
}

 

大数运算之加法

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原文地址：http://www.cnblogs.com/coderAlin/p/4452127.html