[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e8 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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 = ): x(x) { if (x < ) 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 < ) 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 = , v = ;
         while(b) {
             int t = a / b;
             a -= t * b; std::swap(a, b);
             u -= t * v; std::swap(u, v);
         }
         if(u < ) u += MD;
         return u;
     }

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

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

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