国防科学技术大学第忘记叫啥杯了 2015-05-13 21:19 7人阅读 评论(0) 收藏
1594: Factorials
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 207 Solved: 72
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Description
The factorial of an integer N, written N!, is the product of all the integers from 1 through N inclusive. The factorial quickly becomes very large: 13! is too large to store in a 32-bit integer on most computers, and 70! is too large for most floating-point variables. Your task is to find the rightmost non-zero digit of n!. For example, 5! = 1 * 2 * 3 * 4 * 5 = 120, so the rightmost non-zero digit of 5! is 2. Likewise, 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040, so the rightmost non-zero digit of 7! is 4.
Input
This problem includes multiple cases. The first line of input is a integer T represents the number of cases.
For each case, there is a single positive integer N no larger than 4,220.
Output
For each case, output corresponding results separately.
The output of each case is a single line containing but a single digit: the right most non-zero digit of N!.
Sample Input
1
7
Sample Output
4
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int a[100000];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int i,j,k,m;
long long g,t;
double s;
s=0;
for(k=2;k<=n;k++)
s+=log10(k);
m=(int)s+1;//对数累加，至于为啥我也不清楚，感觉挺实用的，求阶乘的位数
for(k=1;k<=m;k++)
{
a[k]=0;
}
a[1]=1;
g=0;
for(k=2;k<=n;k++)
{
for(j=1;j<=m;j++)
{
t=a[j]*k+g;//数组累乘并进位
a[j]=t;
g=t/10;
}
}
for(int i=1;i<=m;i++)
{
if(a[i]!=0)
{
printf("%d\n",a[i]);
break;
}
}
}
return 0;
}
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