hdoj 1002 A + B Problem II

题目描述：

 

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260678    Accepted Submission(s): 50403

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

 

思路：这是一道大数求和问题，用字符数组做就行了。

 

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define M 2000
int a1[M+10];
int a2[M+10];
char s1[M+10];
char s2[M+10];
int main()
{
	int n,k=1,i,j,l1,l2,max,f;
	
	scanf("%d",&n);
	while(n--)
	{
		f=0;
		scanf("%s",s1);
	    scanf("%s",s2);
	    l1=strlen(s1);
	    l2=strlen(s2);
	    max=(l1>l2)?l1:l2;
	    //printf("max=%d\n",max);
	    memset(a1,0,sizeof(a1));
	    memset(a2,0,sizeof(a2));
	    for(i=l1-1,j=0;i>=0;i--)
	    {
	    	a1[j++]=s1[i]-'0';
		}
		for(i=l2-1,j=0;i>=0;i--)
		{
			a2[j++]=s2[i]-'0';
		}
		for(i=0;i<max;i++)
		{
			a1[i]+=a2[i];
			if(a1[i]>=10)
			{
				a1[i]=a1[i]-10;
				a1[i+1]++;
			}
		}
		if(k!=1)
		{
			printf("\n");
		}
		printf("Case %d:\n",k);
		printf("%s + %s = ",s1,s2);
		for(i=max+10;i>=0;i--)
		{
			if(a1[i]==0&&f==0)
			continue;
			else if(a1[i]!=0||f!=0)
			{
				printf("%d",a1[i]);
				f++;
			}
		}
		printf("\n");
        k++;
	}
	
	return 0;
}