 以特定格式输出两个数的和

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 32bit integer. You may assume the length of each integer will not exceed 1000.output
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.Sample Input
2
1 2
112233445566778899 998877665544332211Sample Output
Case 1:
1 + 2 = 3Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
//一看就是ACM题，贴上AC代码
#include<stdio.h>
#include<string.h>
int main()
{
char a[1000],b[1000],c[1001];
int i,j=1,p=0,n,n1,n2;
scanf("%d",&n);
while(n)
{
scanf("%s %s",a,b);
printf("Case %d:\n",j);
printf("%s + %s = ",a,b);
n1=strlen(a)1;
n2=strlen(b)1;
for(i=0;n1>=0n2>=0;i++,n1,n2)
{
if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]'0'+p;}
if(n1>=0&&n2<0){c[i]=a[n1]+p;}
if(n1<0&&n2>=0){c[i]=b[n2]+p;}
p=0;
if(c[i]>'9'){c[i]=c[i]10;p=1;}
}
if(p==1) printf("%d",p);
while(i)
printf("%c",c[i]);
j++;
if(n!=1) printf("\n\n");
else printf("\n");
}
}
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