788.逆序对的数量

倘若q[i]>q[j]，那么在左半边区间i右边的数都大于q[j]
故S[j]=mid-i+1

#include
using namespace std;
const int N = 100010;
int n, m;
int q[N], tmp[N];
long long merge_sort(int q[], int l, int r) {
	if (l >= r) {
		return 0; //TODO
	}
	int mid = (l + r) / 2;
	long long ans = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r); //递归函数遇到return返回结果
	int k = 0, i = l, j = mid + 1;
	while (i <= mid && j <= r) {
		if (q[i] <= q[j]) {    // 这里是小于等于
			tmp[k++] = q[i++];
		} else {              // 注意：只有当q[i]>q[j]才记录逆序对，相等时不计算
			ans += mid - i + 1;
			tmp[k++] = q[j++];
		}
	}
	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[]数组 i的起始位置是l
		q[i] = tmp[j]; //TODO
	}
	return ans;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> q[i];
	}
	long long num = merge_sort(q, 0, n - 1);
	cout << num << endl;

	return 0;
}