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

   
   
   
   

    
    
    
    2
1 2
112233445566778899 998877665544332211
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
1 + 2 = 3

Case 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+10
int 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;
}