利用归并排序求数列中逆序对个数

参考了这篇文章：http://blog.csdn.net/MapReduce/archive/2007/05/20/1618515.aspx
以前自己写的归并排序比较累赘，文章中的代码更有条理，而且计算逆序对个数只需要一行代码，是否对原数列进行排序也只需要决定是否取舍一行代码。
#include <stdio.h>#include <stdlib.h>int ans;void copy(int a[], int b[], int l, int r){int i = l;while (i <= r){a[i] = b[i];i++;}}void merge(int a[], int b[], int l, int m, int r){int i = l, j = m + 1;while (i <= m && j <= r){if (a[i] <= a[j])b[l++] = a[i++];else{b[l++] = a[j++];ans += m - i + 1; // 求逆序对的个数：a[i]到a[m]的每一个数，都会与a[j]构成一个逆序对}}while (i <= m)b[l++] = a[i++];while (j <= r)b[l++] = a[j++];}void merge_sort(int a[], int b[], int l, int r){if (l < r){int m = l + (r - l) / 2;merge_sort(a, b, l, m);merge_sort(a, b, m + 1, r);merge(a, b, l, m, r);copy(a, b, l, r);}}void foo(int a[], int size){int *b = new int[size];merge_sort(a, b, 0, size - 1);delete b;b = NULL;}int main(){int a[] = {6, 5, 4, 3, 2, 1}, i;foo(a, sizeof(a) / sizeof(int));for (i = 0; i < sizeof(a) / sizeof(int); i++)printf("%d ", a[i]);printf("/nans = %d/n", ans);return 0;}