Rqnoj 173 the application of Fish to calculate the number of inverse orders and return the order

The question is in Chinese .. In fact, it is to find the reverse logarithm of a sequence. Use the idea of divide and conquer, and change the order slightly. Code:

 /*  *   Author:        illuz <iilluzen@gmail.com>  *   Blog:          http://blog.csdn.net/hcbbt  *   File:          rqnoj173.cpp  *   Lauguage:      C/C++  *   Create Date:   2013-08-30 09:45:28  *   Descripton:    rqnoj 1173, partitation   */  #include <cstdio>  #define rep(i, n) for (int i = 0; i < (n); i++)  #define repu(i, a, b) for (int i = (a); i < (b); i++)    const int MAXN = 1000100;  int n, cnt, a[MAXN], t[MAXN];    void merge(int l, int r) {      if (r - l <= 1) return;      int m = l + (r - l) / 2;      merge(l, m);      merge(m, r);      int p = l, q = m, i = l;      while (p < m || q < r)          if (q >= r || (p < m && a[p] <= a[q]))              t[i++] = a[p++];          else               t[i++] = a[q++], cnt += m - p;      repu(i, l, r) a[i] = t[i];  }    int main() {      scanf("%d", &n);      rep(i, n) scanf("%d", &a[i]);      cnt = 0;      merge(0, n);      printf("%d\n", cnt);      return 0;  }