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60 lines
1.5 KiB
60 lines
1.5 KiB
#include <iostream>
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#include <string.h>
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#include <stdio.h>
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#include <vector>
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#include <map>
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#include <queue>
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#include <algorithm>
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#include <math.h>
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#include <cstdio>
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using namespace std;
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typedef long long LL;
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const int N = 500010;
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int a[N], q[N], ql;
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int n;
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// 树状数组模板
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typedef long long LL;
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#define lowbit(x) (x & -x)
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int c[N];
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void add(int x, int d) {
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for (int i = x; i < N; i += lowbit(i)) c[i] += d;
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}
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LL sum(int x) {
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LL res = 0;
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for (int i = x; i; i -= lowbit(i)) res += c[i];
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return res;
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}
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// 二分模板 lower_bound (左闭右开)
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int find(int x) {
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return lower_bound(q + 1, q + 1 + ql, x) - q;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ2299.in", "r", stdin);
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#endif
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while (~scanf("%d", &n) && n) {
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// 清空树状数组
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memset(c, 0, sizeof c);
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// 读入原始数组
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for (int i = 1; i <= n; i++) scanf("%d", &a[i]), q[i] = a[i];
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// 由小到大排序+离散化
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sort(q + 1, q + 1 + n);
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ql = 1;
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for (int i = 2; i <= n; i++)
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if (q[ql] != q[i]) q[++ql] = q[i];
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// 利用树状数组动态维护个数，动态获取前缀和（权值）
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LL ans = 0;
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for (int i = 1; i <= n; i++) {
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// 在我前面进来，比我大的有多少个？
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ans += sum(ql) - sum(find(a[i]));
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add(find(a[i]), 1);
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}
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printf("%lld\n", ans);
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}
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return 0;
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} |