[hdu5225]逆序对统计

题目：给定一个1到n的排列，求字典序小于这个排列的所有排列的逆序对数之和。

思路：既然是求字典序小于这个排列的，不妨将排列根据和它前k位相同来分类，然后枚举第k+1位的数（小于原序列第k+1位的数），假设逆序对的位置为(x,y)，对于1<=x<k+1，x<y<=k+1和1<=x<=k+1,k+2<=y<=n的答案是容易计算出来的，对于k+2<=x<n,x<y<=n的答案则可以通过dp来计算由于剩余的数已没有大小意义了，假设剩余p个不同的数，则p个数的全排列产生的逆序对总数与1-p这p个数产生的全排列的逆序对总数是相同的，所以可以令dp[n]表示n个数产生的全排列的逆序对总数，则dp[n] =dp[n-1]*n+C(n,2)*(n-1)!。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef unsigned int uint;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e8 + 17;
const int md = 1e9 + 7;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

template<int mod>
struct ModInt {
    const static int MD = mod;
    int x;
    ModInt(int x = 0): x(x) { if (x < 0) x += mod; }
    int get() { return x; }

    ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
    ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
    ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
    ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

    ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
    ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
    ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
    ModInt operator /= (const ModInt &that) { *this = *this / that; }

    ModInt inverse() const {
        int a = x, b = MD, u = 1, v = 0;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += MD;
        return u;
    }

};
typedef ModInt<md> mint;
mint dp[107], fact[107];
int n, a[107];
bool vis[107];

void init() {
    fact[0] = 1;
    rep_up1(i, 102) {
        dp[i] = dp[i - 1] * i + fact[i - 1] * i * (i - 1) / 2;
        fact[i] = fact[i - 1] * i;
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    init();
    while (cin >> n) {
        rep_up0(i, n) sint(a[i]);
        mem0(vis);
        mint ans = 0;
        rep_up0(i, n) {
            int c = 0;
            rep_up0(j, i) {
                for (int k = j + 1; k < i; k ++) {
                    if (a[j] > a[k]) c ++;
                }
            }
            rep_up1(j, a[i] - 1) {
                if (!vis[j]) {
                    ans += fact[n - i - 1] * c;
                    rep_up0(k, i) {
                        if (a[k] > j) ans += fact[n - i - 1];
                    }
                    int sum = 0;
                    vis[j] = true;
                    rep_up1(k, n) {
                        if (!vis[k]) sum ++;
                        else ans += fact[n - i - 1] * sum;
                    }
                    ans += dp[n - i - 1];
                    vis[j] =false;
                }
            }
            vis[a[i]] = true;
        }
        cout << ans.get() << endl;
    }
    return 0;
}