杭电 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260805    Accepted Submission(s): 50438

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

 

总算过了，表示细节害死人啊！

#include<stdio.h>#include<string.h>#define max 1000+10int a[max],b[max];char str1[max],str2[max];int main(){int m;int k=1;scanf("%d",&m);while(m--){scanf("%s %s",str1,str2);memset(a,0,sizeof(a));memset(b,0,sizeof(b));int i,j;for(i=0,j=strlen(str1)-1;i<strlen(str1);i++){a[j--]=str1[i]-'0';}for(i=0,j=strlen(str2)-1;i<strlen(str2);i++){b[j--]=str2[i]-'0';}for(i=0;i<max;i++){a[i]+=b[i];if(a[i]>=10){a[i]-=10;a[i+1]+=1;}}printf("Case %d:\n",k++);printf("%s + %s = ",str1,str2);    for(i=max-1;(i>=0)&&(a[i]==0);i--);if(i>=0){for(;i>=0;i--){printf("%d",a[i]);}}else printf("0");if(m!=0) printf("\n\n");else printf("\n");}return 0;}