> #include<stdio.h>
> #include<math.h>
> int fn(long num)
> {
> 	if(num<=9)
> 		return num;
> 	else if(num<=189)
> 	{
> 		num-=10;
> 		if(num%2==0)
> 			return num/20+1;
> 		else return (num%20-1)/2%10;
> 	}
> 	else if(num<=2889)
> 	{
> 		num-=190;
> 		if(num%3==0)
> 			return num/300+1;
> 		else if((num-1)%3==0)
> 			return (num%300-1)/30;
> 		else if((num-2)%3==0)
> 			return (num%300-2)%30/3;
> 	}
> 	else if(num<=38889)
> 	{
> 		num-=2890;
> 		if(num%4==0)
> 			return num/4000+1;
> 		else if((num-1)%4==0)
> 			return (num%4000-1)/400;
> 		else if((num-2)%4==0)
> 			return (num%4000-2)%400/40;
> 		else if((num-3)%4==0)
> 			return (num%4000-3)%40/4;
> 	}
> 	else
> 	{
> 		num-=38890;
> 		if(num%5==0)
> 			return num/50000+1;
> 		else if((num-1)%5==0)
> 			return (num%50000-1)/5000;
> 		else if((num-2)%5==0)
> 			return (num%50000-2)%5000/500;
> 		else if((num-3)%5==0)
> 			return (num%50000-3)%500/50;
> 		else if((num-4)%5==0)
> 			return (num%50000-4)%50/5;
> 	}
> }
> long clj(int a1,unsigned long a,long b,int c)
> {
> 	long low=1,high=b,mid;
> 	unsigned long sum,sum1;
> 	while(low<=high)
> 	{
> 		mid=(low+high)/2;
> 		if(mid<1000)
> 		{
> 			sum=(2*a1+c*(mid-1))*mid/2;
> 			sum1=(2*a1+c*(mid-2))*(mid-1)/2;
> 		}
> 		else
> 		{
> 			sum=((2*a1+c*(mid-1))/2)*mid;
> 			sum1=((2*a1+c*(mid-2))/2)*(mid-1);
> 		}
> 		if(a<=sum&&a>sum1)
> 			return (a-sum1-1)%(a1+c*(mid-1))+1;
> 		else if(sum>a)
> 			high=mid-1;
> 		else if(sum<a)
> 			low=mid+1;
> 	}
> }
> int main()
> {
> 	int t,ans;
> 	unsigned long n;
> 	long num;
> 	scanf("%d",&t);
> 	while(t--)
> 	{
> 		scanf("%lu",&n);
> 		if(n<=45)
> 			num=clj(1,n,9,1);
> 		else if(n<=9045)
> 			num=clj(11,n-45,90,2);
> 		else if(n<=1395495)
> 			num=clj(192,n-9045,900,3);
> 		else if(n<=189414495)
> 			num=clj(2893,n-1395495,9000,4);
> 		else num=clj(38894,n-189414495,90000,5);
> 		ans=fn(num);
> 		printf("%d\n",ans);
> 	}
> }

• 终于过了,主要是精度问题 (0) windbells 2004-07-09 18:47:30