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A + B Problem II

2
1 2
112233445566778899 998877665544332211

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include<stdio.h>
#include<string.h>
#define MAX 1005
int a1[MAX];
int a2[MAX];
char s1[MAX];
char s2[MAX];
int main()
{
	int n,t=1;
	scanf("%d",&n);
	while(n--)
	{
		int i,j,length1,length2;
		scanf("%s",s1);
		scanf("%s",s2);
		memset(a1,0,sizeof(a1));
		memset(a2,0,sizeof(a2));
		length1=strlen(s1);
		for(j=0,i=length1-1;i>=0;i--)
		a1[j++]=s1[i]-'0';
		length2=strlen(s2);
		for(j=0,i=length2-1;i>=0;i--)
		a2[j++]=s2[i]-'0';
		for(i=0;i<MAX;i++)
		{
			a1[i]=a1[i]+a2[i];
			if(a1[i]>=10)
			{a1[i]-=10;a1[i+1]++;}
		}
		printf("Case %d:\n",t++);
		printf("%s + %s = ",s1,s2);
		for(i=MAX;(i>=0)&&(a1[i]==0);i--);
		if(i>=0)
		{
			for(;i>=0;i--)
			printf("%d",a1[i]);
			printf("\n");
		}
		else
		printf("0\n");
		if(n>0)
		printf("\n"); 
	}
	return 0;	
}

HDU 1002 A + B Problem II,布布扣,bubuko.com

HDU 1002 A + B Problem II

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原文地址：http://blog.csdn.net/qq_16767427/article/details/38107971