AcWing 788. 逆序对的数量
原题链接
简单
作者:
跟着灿哥学切菜
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2021-01-13 11:48:05
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所有人可见
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阅读 270
#include <iostream>
using namespace std;
const int N = 100010;
int q[N], tmp[N];
typedef long long LL;
LL merge_sort(int q[], int l, int r) {
if (l >= r) return 0;
int mid = l + r >> 1, i = l, j = mid + 1;
LL res = merge_sort(q, i, mid) + merge_sort(q, j, r);
int k = 0;
while (i <= mid && j <= r) {
//此时如果是左部分的某一个值小于等于右边数组中的某一个值, 此时不构成逆序对,直接装箱
if (q[i] <= q[j]) tmp[k ++] = q[i ++];
else {
tmp[k ++] = q[j ++];
res += mid - i + 1;
}
}
while (i <= mid) tmp[k ++] = q[i ++];
while (j <= r) tmp[k ++] = q[j ++];
for (int i = l, j = 0; i <= r; i ++, j ++) q[i] = tmp[j];
return res;
}
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++) scanf("%d", &q[i]);
cout << merge_sort(q, 0, n - 1) << endl;
return 0;
}