hdu 4196 Remoteland

转自：

http://blog.youkuaiyun.com/acm_cxlove/article/details/7422265

Remoteland

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 399    Accepted Submission(s): 156

Problem Description

In the Republic of Remoteland, the people celebrate their independence day every year. However, as it was a long long time ago, nobody can remember when it was exactly. The only thing people can remember is that today, the number of days elapsed since their independence (D) is a perfect square, and moreover it is the largest possible such number one can form as a product of distinct numbers less than or equal to n.
As the years in Remoteland have 1,000,000,007 days, their citizens just need D modulohttp://blog.youkuaiyun.com/acm_cxlove/article/details/7422265Note that they are interested in the largest D, not in the largest D modulo 1,000,000,007.

 

Input

Every test case is described by a single line with an integer n, (1<=n<=10,000, 000). The input ends with a line containing 0.

 

Output

For each test case, output the number of days ago the Republic became independent, modulo 1,000,000,007, one per line.

 

Sample Input

  

   4
9348095
6297540
0
  

 

Sample Output

  

   4
177582252
644064736
  

 

Source

SWERC 2011

题目：用不大于n内的所有数去组成一个尽可能大的完全平方数。

http://acm.hdu.edu.cn/showproblem.php?pid=4196

完全平方数，显然是所有的素因子的个数都是偶数，便取N！的所有素因子的个数，便 有了初步想法，算出N！的什么素因子个数，奇数的就舍去一个，偶数的全要，然后再全部乘起来，可是因为规模很大，即使快速幂乘也是会超时。

于是考虑把N！除掉那些奇数个因子的乘积，便是求a=c/b，由于是取模的，直接除必然不行。

考虑c%mod=(a%mod)*(b%mod);令A=a%mod; B=b%mod;C=c%mod;

则A=a%mod=(a*1)%mod=a%mod*1%mod=（a%mod）*（b^(mod-1)）%mod      --------因为(b^(mod-1))%mod=1,费马小定理。

=(a*b)%mod*(b^(mod-2))%mod=(c%mod)*(b%mod)^(mod-2)%mod=C*B^(mod-2)

利用这个，就可以求出c/b的结果。

则题目求解的步骤便是：

1、打素数表

2、计算N！中含有素因子个数，如果为奇数就保留这个素因子的乘积，以便最后相除

3、求出除法结果

其实求解逆原的是重点，任何数 A ^ A^*(mod-2) == 1 % (mod)  mod=1,000,000,007, A 的摸上mod 逆原就是A^*(mod-2) ，任何数 除 A 相当于 *  A^*(mod-2)

#include<iostream>
#include<cstdio>
#include<cmath>
#define N 10000000
#define MOD 1000000007
#define LL long long
using namespace std;
bool flag[N+5]={0};
int prime[1000000],cnt=0;
void Prime(){
	for(int i=2;i<sqrt(N+1.0);i++){
		if(flag[i])
			continue;
		for(int j=2;j*i<=N;j++)
			flag[i*j]=true;
	}
	for(int i=2;i<=N;i++)
		if(!flag[i])
			prime[cnt++]=i;
}
int getsum(int n,int p){
	int sum=0;
	while(n){
		sum+=n/p;
		n/=p;
	}
	return sum;
}
LL PowMod(LL a, LL b){
	LL ans=1;
	while(b){
		if(b&1)
			ans=(ans*a)%MOD;
		a=(a*a)%MOD;
		b>>=1;
	}
	return ans;
}
LL fac[N+5];
int a;
int main(){
	int n;
	fac[1]=1;
	for(int i=2;i<=N;i++)
		fac[i]=(fac[i-1]*i)%MOD;
	Prime();
	while(scanf("%d",&n)!=EOF&&n){
		LL ret=1;
		for(int i=0;i<cnt&&prime[i]<=n;i++){
			a=getsum(n,prime[i]);
			if(a&1)
				ret=(ret*prime[i])%MOD;
		}
		printf("%I64d\n",(fac[n]*PowMod(ret,MOD-2))%MOD);
	}
	return 0;
}