技術標籤：經典例題c語言

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

Sample Input

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

這道題的陷阱就是格式輸出

#include<stdio.h>
#include<string.h>
long long a[1050],a1[1050],sum[1050];
char s[1050],s1[1050];
long long i,j,l,l1,t,n=1;
int main()
{
	scanf
 
("%d",&t);
    for(n=1;n<=t;n++)
    {
    	scanf("%s %s",s,s1);
        l=strlen(s);l1=strlen(s1);
        for(i=0;i<=1040;i++)
        {
            a[i]=0;a1[i]=0;sum[i]=0;
        }
        for(i=l-1,j=1;i>=0;i--,j++)
        a[j]=s[i]-'0';
        for(i=l1-1,j=1;i>=0;i--,j++)
        a1[j]=s1[i]-'0';
        for(i=0;i<=1010;i++)
        {
            sum[i]+=a1[i]+a[i];
            if(sum[i]>=10)
            {
                sum[i+1]++;
                sum[i]=sum[i]%10;
            }
        }
        for(i=1010;i>=0;i--)
        {
            if(sum[i]!=0)
            break;
        }
        printf("Case %d:\n",n);
        printf("%s + %s = ",s,s1);
        if(i==-1)
		printf("0");
        for(j=i;j>0;j--)
        printf("%lld",sum[j]);
        if(t!=n)
        printf("\n\n");
        else printf("\n");
    }
    return 0;
}