C语言之大数加法

待我称王封你为后i 2022-06-05 02:51 212阅读 0赞

## 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][Link 1]

#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;
    }

[Link 1]: http://59.69.128.203/JudgeOnline/search_result.php?source=%E7%BB%8F%E5%85%B8%E9%A2%98%E7%9B%AE