poj 2299 求逆序对

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经典题目：给定一个序列，求其中的逆序对

在归并排序(merge sort)的基础上统计，时间复杂度O(nlogn)

注意使用long long

#include <iostream>#include <vector>#include <map>#include <list>#include <set>#include <deque>#include <stack>#include <queue>#include <algorithm>#include <cmath>#include <cctype>#include <cstdio>#include <iomanip>#include <cmath>#include <cstdio>#include <iostream>#include <string>#include <sstream>#include <cstring>#include <queue>using namespace std;///宏定义const int  INF = 10000000;const int MAXN = 500100;const int maxn = MAXN;///全局变量 和 函数//int T;int n;long long A[maxn], T[maxn];long long cnt;long long quick_sort(long long *A, long long x, long long y, long long *T){    long long cnt = 0;    if(y - x > 1)    {        long long m = x + (y - x) / 2;        long long p = x, q = m, i = x;        long long leftt = quick_sort(A, x, m, T);        long long rigtt = quick_sort(A, m, y, T);        while(p < m || q < y)        {            if(q >= y || (p < m && A[p] <= A[q]))            {                T[i++] = A[p++];            }            else            {                T[i++] = A[q++];                cnt += m - p;       //添加这一句即可，其它同归并排序            }        }        for(i = x; i < y; i++)            A[i] = T[i];        cnt += leftt;        cnt += rigtt;    }    return cnt;}int main(){///变量定义int i, j;    while(scanf("%d", &n) == 1 && n)    {        for(i = 0; i < n; i++)            scanf("%lld", &A[i]);        printf("%lld\n", quick_sort(A, 0, n, T));    }///结束return 0;}