Hdoj 1002

问题描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

输入描述

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.

输出描述

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. Output a blank line between two test cases.

样例输入

2

1 2

112233445566778899 998877665544332211

样例输出

Case 1:

1 + 2 = 3

 

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

思路

简单的大数问题

代码

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

int main()
{
    int n, j;
    scanf("%d", &n);
    for(j = 1; j <= n; j++) 
    {
        char ch1[1001], ch2[1001];
        scanf("%s", ch1);
        int s1 = strlen(ch1), i;
        int a[1001] = {0}, b[1001] = {0};
        for(i = s1-1; i >= 0; i--) { a[s1-i] = ch1[i] - '0'; }
        scanf("%s", ch2);
        int s2 = strlen(ch2);
        for(i = s2-1; i >= 0; i--) { b[s2-i] = ch2[i] - '0'; }
        printf("Case %d:
%s + %s = ", j, ch1, ch2);
        int s = (s1>s2?s1:s2), temp=0;
        for(i = 1 ; i <= s; i++)
        {
            a[i] += b[i] + temp;
            temp = a[i] / 10;
            a[i] %= 10;
        }
        if(temp != 0) printf("%d",temp);
        for(i=s;i>0;i--) printf("%d",a[i]);
        if(j < n) printf("
");
        printf("
");
    }
}