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就是这样

Posted by zhucheng at 2005-10-05 17:00:13 on Problem 1999
In Reply To:请教牛人,不知道是不是想法不对。 Posted by:wr at 2005-10-05 16:56:50
> 我的想法是打牌的人每种牌有且只有一张,没有顺序关系。判断是不是能够把其他两人的消掉。判断条件是在配对的牌间连线,看有没有相交的.
> 有AC的懒得说就把code站内发一下,谢了。
> 
> #include <stdio.h>
> #include <string.h>
> #define max 100008
> long a[max],b[max];
> char pk[max*10]; 
> int main(){
> 	long z,i,m,n,p,l;
> 	for (scanf("%ld",&z);z>0;z--) {
> 		for (i=1;i<=10*max;i++) pk[i]=0;
> 		scanf("%ld%ld",&m,&n);  p=1; l=-1;
> 		for (i=1;i<=m;i++) { 
> 			scanf("%ld",&a[i]);	if (a[i]>=10*max)for (;;);
> 			pk[a[i]]=1; 
> 		}
> 		for (i=1;i<=n;i++) {
> 			scanf("%ld",&b[i]);	
> 			if (p&&b[i]<10*max&&pk[b[i]]==1) {
> 				if (l==-1||b[i]>l) l=b[i];
> 				else p=0;
> 			}
> 		}
> 		if (p==1) printf("Hrac ma sanci vyhrat.\n");
> 		else printf("Spatne usporadani.\n");
> 	}
> 	return 0;
> }

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