hust 1413 Permutation Design

题目描述

       xiaoY is crazy about solving equations! But since teacher HH tells him that when an equation’s degree is bigger than 3, there are no analytic solutions, so he feels very upset! With a few days of trial, he truly trusts his teacher’s words and he focuses on another interesting problem. What is the “Permutation function group” like!

       A “Permutation function group” means the coefficient of the function is a permutation of 1 ~N and the format is

        F(x) = a1*x^ (1^3) + a2*x^ (2^3) … + an*x^ (N^3);

He is interested about what the F(x) is like when a1...aN form the Kth permutation of all;

As you see, when N=4:

       S (1):    1 2 3 4

       S (2):     1 2 4 3

       S (3):     1 3 2 4

       …

      S (N)      4 3 2 1

Now give you N, k and x, you just need to tell xiaoY F(x) mod 1000000007 (in case F(x) too large! We module 1000000007)

输入

Multiple cases.

N, K, X (0<N<=1000, 0<K<=3000, x<10^9)

0 0 0 indicates the end of input!

输出

F(x) mod 1000000007

样例输入

2 2 2
0 0 0

样例输出

260

提示

2 * 2^(1^3) + 1 * 2^(2^3) = 4 + 256 = 260

这个题考了两个知识点

1，怎么求下一个排列

2，当然是快速幂了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;

const long long mod=1000000007;
long long a[1001];
void init()
{
    for (long long i=1;i<=1000;i++)
    a[i]=i*i*i;
}

int f[1001];

long long find(long long n,long long x)
{
    long long e=1,tmp=x;
    while(n)
    {
        if (n & 1)
        e=(e*tmp)%mod;
        tmp=(tmp*tmp)%mod;
        n >>=1;
    }
    return e%mod;
}

int main()
{
    init();
    long long n,k,x,ans;
    while (scanf("%lld%lld%lld",&n,&k,&x)!=EOF && !(x==0 && n==0 && k==0))
    {
        for (int i=1;i<=n;i++) f[i]=i;
        k--;
        while(k--)
        {
            int m=next_permutation(f+1,f+n+1);
        }
        ans=0;
        for (int i=1;i<=n;i++)
        {
            ans+=(f[i]*find(a[i],x))%mod;
        }
        printf("%lld\n",ans%mod);
    }
    return 0;
}