#include <stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
	int n,i,j,len,k,h,fac,need,sum;
	char id[8];
	scanf("%d",&n);	
	for(i=1;i<=n;i++)
	{
		scanf("%s",id);
		len=strlen(id);
		sum=0;
		for(j=len-1;j>=0;j--)
		{
			if(id[j]!='?')
			{
				if((len-j)%3==1)
					sum+=(id[j]-'0')*9;
				else if((len-j)%3==2)
					sum+=(id[j]-'0')*3;
				else
					sum+=(id[j]-'0')*7;
			}
			else
				k=j;
		}
		need=10-sum%10;        \\要加数的个位数字 
		printf("Scenario #%d:\n",i);
		if((len-k)%3==1)
			fac=9;
		else if((len-k)%3==2)
			fac=3;
		else 
	    	fac=7;
		for(h=1;h<=9;h++)
			if((h*fac)%10==need)
				break;
		id[k]=h+'0';
		printf("%s\n\n",id);	
	}
	system("pause");
	return 0;
}

• Re:超时，大牛帮帮忙，简化一下啊 (821) 10074182 2009-08-04 16:11:32