【HDU

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 modulo 1,000,000,007. Note 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
题意：从[1,n]之间选择若干个数字，使其为平方数，现在求最大的这个平方数。

分析： 从算数基本定理来考虑， 如果一个数字是平方数那么其所有质因子的指数都是偶数，从这一点来考虑，[1,n]之间最大的我们首选n! ， 然后我们只需要枚举n!的所有的质因子，看其质因子的指数，如果是偶数，我们不用管，但是如果是奇数，我们就要除去一个当前质因子(因为如果有当前的这个质因子的话，那么[1,n]之间一定有一个数字正好等于当前的这个质因子，所以我们除掉一个，就相当于将这个单独的数去掉，不会影响什么)。

代码

 #include<algorithm> #include<stdio.h> #define LL long long  using namespace std;const int MAXN = 1e7+11;const int MAXM = 1e7+11;const int mod =  1e9+7;const int inf = 0x3f3f3f3f;int prm[MAXN+2],sz; bool su[MAXN+2]; int fac[MAXM];  void init(){    su[0]=su[1]=true;    for(int i=2;i<=MAXN;i++){        if(!su[i])  prm[++sz]=i;        for(int j=1;j<=sz;j++){            int t=i*prm[j];            if(t>MAXN) break;            su[t]=true;            if(i%prm[j]==0) break;        }    }    fac[0]=fac[1]=1;     for(int i=2;i<=MAXM;i++){        fac[i]=(LL)fac[i-1]*i%mod;    }}LL power(LL a,LL b,LL c){    LL s=1,base=a%c;    while(b){        if(b&1) s=s*base%c;        base=base*base%c;        b>>=1;    }    return s;}LL inv(LL a){  // 费马小定理求逆元    return power(a,mod-2,mod);}void solve(LL n){    LL ans=fac[n]; LL temp=1;    for(int i=1;i<=sz;i++){        LL cnt=0; LL t=n;        if(n/prm[i]){            while(t){                cnt+=t/prm[i];                t/=prm[i];            }             if(cnt&1) temp=temp*prm[i]%mod;  // 将所有要除的先都存起来，最后求一下逆元， 一开始这里直接就求逆元了，无限TLE。        }else break;     }    ans=ans*inv(temp)%mod; // 只要这里求一次逆元就行了     printf("%lld\n",ans);}int main(){     LL n; init();     //printf("%d\n",sz);     while(scanf("%lld",&n)&&n){          solve(n);    }    return 0;}