杭电oj 1002

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260218 Accepted Submission(s): 50305

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.

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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

#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>=0||n2>=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");}n--;}}

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