#include<stdio.h>
#include<string.h>
//FILE *fa;
int s[66]={0},a,b,c[66]={0};
int i,t,k,i1,i2,i3,i4,sum;
long o[1000][5][5],l;
int tie[1000][5][5];
	ex[1000];
int main()
{
//	fa=fopen("l.in","r");
	while(scanf("%d",&s[1])!=EOF)
	{
		a=1;
		while(s[a])
		{
			a++;
			scanf("%d",&s[a]);
		}
		a--;
		b=0;
		scanf("%d",&c[0]);
		while(c[b])
		{
			b++;
			scanf("%d",&c[b]);
		}
		memset(o,0,sizeof(o));
		memset(tie,0,sizeof(tie));
		memset(ex,0,sizeof(ex));
		for(i=a;i>=0;i--)
			for(i1=i;i1>=0;i1--)
				for(i2=i1;i2>=0;i2--)
					for(i3=i2;i3>=0;i3--)
					{
						t=0;
						sum=s[i]+s[i1]+s[i2]+s[i3];
						if(i) t++;if(i2)t++;
						if(i1)t++;if(i3)t++;
						k=t;
						if(i==i1&&i)  k--;
						if(i1==i2&&i1)k--;
						if(i2==i3&&i2)k--;
						l=((i*100+i1)*100+i2)*100+i3;
						if(o[sum][k][t])
						{
							if(o[sum][k][t]/1000000==i)
								tie[sum][k][t]=1;
							else
							if(o[sum][k][t]/1000000<i)
							{
								tie[sum][k][t]=0;
								o[sum][k][t]=l;
							}
						}
						else
						if(k>0)
						{
							o[sum][k][t]=l;
							ex[sum]=1;
						}
					}
		for(i=0;i<b;i++)
			if(!ex[c[i]])
				printf("%d ---- none\n",c[i]);
			else
			for(i1=a;i1>0;i1--)
				for(i2=i1;i2<5;i2++)
					if(tie[c[i]][i1][i2])
					{
						printf("%d (%d): tie\n",c[i],i1);
						i1=0;i2=5;
					}
					else
					if(o[c[i]][i1][i2])
					{
						printf("%d (%d):",c[i],i1);
						t=o[c[i]][i1][i2];
						for(i3=4;i3>0;i3--)
						{
							k=t%100;
							t=t/100;
							if(k)
								printf(" %d",s[k]);
						}
						printf("\n");
						i1=0;i2=5;
					}			
	}
	return 0;
}

• Re:枚举~~~- - 附代码 (69) graint 2012-03-30 10:27:48
• 这必须具备以下两个条件 (74) HeadHunter 2013-05-12 11:00:10
• Re:枚举~~~- - 附代码 (1863) sd15w2015012893 2017-04-11 08:28:18