0
点赞
收藏
分享

微信扫一扫

HDU5139:Formula(找规律+离线处理)

http://acm.hdu.edu.cn/showproblem.php?pid=5139

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

官方题解:

找规律
f(1)=1
f(2)=1*1*2=(1)*(1*2)=1!*2!
f(3)=1*1*1*2*2*3=(1)*(1*2)*(1*2*3)=1!*2!*3!

式子可以简化为 f(n)=∏i=1n(n!)%MOD,直接打表不行,会超内存,可以对数据进行离线处理。排好序之后从小到大暴力。ClogC+10000000 ,C为case数目。

题目解析:以前根本不知道题目可以这么做,又学了一样新东西,离线处理。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <queue>
#include <map>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
const int mod=1000000007;
using namespace std;
int n,tt;
struct node
{
    int id,x,sum;
} q[100010];
int cmp1(const void *a,const void *b)
{
    struct node *aa=(struct node *)a;
    struct node *bb=(struct node *)b;
    return aa->x-bb->x;
}
int cmp2(const void *a,const void *b)
{
    struct node *aa=(struct node *)a;
    struct node *bb=(struct node *)b;
    return aa->id-bb->id;
}
int main()
{
    tt=0;
    __int64 s=1,s2=1;
    while(scanf("%d",&n)!=EOF)
    {
        q[tt].id=tt;
        q[tt++].x=n;
    }
    qsort(q,tt,sizeof(q[0]),cmp1);
    for(int i=0,j=2; i<tt; i++)
    {
        for(; j<=q[i].x; j++)
        {
            s=(s*j)%mod;
            s2=(s*s2)%mod;
        }
        q[i].sum=s2;
    }
    qsort(q,tt,sizeof(q[0]),cmp2);
    for(int i=0; i<tt; i++)
        printf("%d\n",q[i].sum);
    return 0;
}

 


 



举报

相关推荐

0 条评论