HDU 5139 Formula

最新推荐文章于 2019-11-24 22:08:39 发布

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Formula

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 384 Accepted Submission(s): 160

Problem Description

f(n)=(∏i=1nin−i+1)%1000000007
You are expected to write a program to calculate f(n) when a certain n is given.

Input

Multi test cases (about 100000), every case contains an integer n in a single line.
Please process to the end of file.

[Technical Specification]

1≤n≤10000000

Output

For each n，output f(n) in a single line.

Sample Input

  
   2
100

Sample Output

  
   2
148277692

Source

BestCoder Round #21

题目蛮好理解意思的

找找规律

f(1)=1^1;

f(2)=1^2*2^1=1!*2!;

f(3)=1^3*2^2*3^1=1!*2!*3!;

于是就想到了预处理。。来了一发，结果内存爆了。。经过百度题解，找到了离线处理的方法。。先排个序，然后把得到的按照从小到大的输入暴力一遍就搞定了

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
#define mod 1000000007
using namespace std;
int s[10000010],a[10000010],f[10000010];
int main()
{
	int n=-1,i,j;
	while( ~scanf("%d",&a[++n]))s[n]=a[n];        
    sort(a,a+n) ;          
    long long Sum=1,sum=1,k=1;  
    for(i=0;i<=n-1;i++) 
	{ 
        for(j=k;j<=a[i];j++)
		{  
            sum=sum*j%mod;  
        	Sum=Sum*sum%mod;  
	 	}  
	 	k=a[i]+1;
	 	f[a[i]]=Sum;
	}
	for(i=0;i<n;i++)
	{
		printf("%d\n",f[s[i]]);
	}
	
}