49 lines
1.0 KiB
C++
49 lines
1.0 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 3e5 + 10;
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int n;
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int q[N];
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int t[N];
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LL ans;
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/*
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6
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0 1 2 3 4 5
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答案：15
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*/
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// 归并排序，附带求顺序对的数量
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void merge_sort(int q[], int l, int r) {
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if (l >= r)
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return;
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int mid = (l + r) >> 1;
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merge_sort(q, l, mid), merge_sort(q, mid + 1, r);
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int i = l;
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int j = mid + 1;
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int k = 0;
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while (i <= mid && j <= r)
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if (q[i] <= q[j]) {
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t[k++] = q[i++];
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// 顺序对，原理：当q[i]<=q[j]时，q[i]与q[j]之后的所有数都构成顺序对
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ans += r - j + 1;
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} else
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t[k++] = q[j++];
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while (i <= mid)
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t[k++] = q[i++];
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while (j <= r)
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t[k++] = q[j++];
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for (i = l, j = 0; i <= r; i++, j++)
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q[i] = t[j];
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}
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++)
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cin >> q[i];
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// 根据题目提示，使用归并排序，并在归并排序的过程中计算顺序对的数量
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merge_sort(q, 1, n);
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printf("%lld", ans);
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return 0;
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}
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