hdu2110Crisis of HDU-母函数

hdu2110Crisis of HDU

0<pi,mi<10 这个条件貌似不对，用p[i] m[i]分别储存才过（之前是模板的方法）

#include<iostream>
using namespace std;

int c1[4000],c2[4000],X[101],P[101];

int main()
{
	int T,n,m,k,a,b,i,j,num,sum;
	
	while(scanf("%d",&n),n)
	{
		memset(X,0,sizeof(X));
		for(sum=0,i=1;i<=n;i++)
		{
			scanf("%d%d",&P[i],&X[i]);
			sum+=P[i]*X[i];
		}
		
		if(sum%3)
		{
			printf("sorry\n");
			continue;
		}
		sum/=3;
		memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));
		
		for(i=1;i<=n;i++)
			if(X[i])
			{
				for(num=0;num<=X[i];num++)
					if(num*P[i]<=sum)	c1[num*P[i]]=1;
					else	break;
				break;
			}
		
		for(i=i+1;i<=n;i++)
		{
			if(X[i])
			{
				for(j=0;j<=sum;j++)
					if(c1[j])
						for(num=0;num<=X[i];num++)
							if(num*P[i]+j<=sum)	c2[num*P[i]+j]+=c1[j];
							else break;
				for(j=1;j<=sum;j++)
				{
					c1[j]=c2[j]%10000;
					c2[j]=0;
				}
			}
		}
		if(c1[sum]==0)	printf("sorry\n");
		else	printf("%d\n",c1[sum]%10000);
	}
	return 0;
}