HDU 5225 Tom and permutation

最新推荐文章于 2015-06-13 12:45:59 发布

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最新推荐文章于 2015-06-13 12:45:59 发布

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分类专栏： ----数位dp HDU 文章标签： hdu

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本文介绍了一个算法问题：给定一个1到n的排列，求所有比该排列字典序小的排列中各排列的逆序数之和，并提供了一个具体的C++实现方案，包括预处理阶乘和逆序数计算。

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Problem Description

Tom has learned how to calculate the number of inversions in a permutation of n distinct objects by coding, his teacher gives him a problem:
Give you a permutation of n distinct integer from 1 to n, there are many permutations of 1-n is smaller than the given permutation on dictionary order, and for each of them, you can calculate the number of inversions by coding. You need to find out sum total of them.
Tom doesn't know how to do, he wants you to help him.
Because the number may be very large, output the answer to the problem modulo

109+7.

Input

Multi test cases(about 20). For each case, the first line contains a positive integer n, the second line contains n integers, it's a permutation of 1-n.

n≤100

Output

For each case, print one line, the answer to the problem modulo

109+7.

Sample Input

Sample Output

1
75
数位dp
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 105;
const int base = 1000000007;
int n, a[maxn], u[maxn];
long long f[maxn], p[maxn], ans, tot, sum;

void get()
{
    p[0] = 1;
    for (int i = 1; i <= 100; i++)
    {
        f[i] = (i * f[i - 1] + p[i - 1] * i * (i - 1) / 2) % base;
        p[i] = (p[i - 1] * i) % base;
    }
}

int main()
{
    get();
    while (scanf("%d", &n) != EOF)
    {
        ans = tot = 0;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]), u[i] = 1;
        for (int i = 1; i <= n; i++)
        {
            sum = 0;
            for (int j = 1; j < a[i]; j++)
            if (u[j])
            {
                (ans += f[n - i] + p[n - i] * (tot+sum)) %= base;
                sum++;
            }

            tot += sum;
            u[a[i]] = 0;
        }
        cout << ans << endl;
    }
    return 0;
}