HDU 5196 DZY Loves Inversions(树状数组，二分)

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这题之前CC做过类似的，思路如官方题解。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 100005;typedef long long ll;int n, m;ll q;struct Hash {    int v, id;    void read(int id) {        scanf("%d", &v);        this->id = id;    }} h[N];bool cmp(Hash a, Hash b) {    return a.v < b.v;}int a[N];int bit[N], l[N];inline int lowbit(int x) {return (x&(-x));}void add(int x, int v) {    while (x <= n) {        bit[x] += v;        x += lowbit(x);    }}int get(int x) {    int ans = 0;    while (x) {        ans += bit[x];        x -= lowbit(x);    }    return ans;}ll sum[N], sum2[N];ll ans[N];int X[N], Y[N];void gao(ll q, int bo) {    if (q < 0) return;    memset(bit, 0, sizeof(bit));    int r = 0;    ll tot = 0;    for (int i = 1; i <= n; i++) {        while (r <= n && tot <= q) {            r++;            if (r > n) break;            tot += (r - i) - get(a[r]);            add(a[r], 1);        }        l[i] = r - 1;        add(a[i], -1);        tot -= get(a[i] - 1);    }    sum[n + 1] = 0;    for (int i = n; i >= 1; i--)        sum[i] = sum[i + 1] + (l[i] - i + 1);    int x, y;    for (int i = 0; i < m; i++) {        int x = X[i], y = Y[i];        int tmp = lower_bound(l + x, l + y + 1, y) - l - 1;        ans[i] += bo * (sum[x] - sum[tmp + 1] + sum2[y - tmp]);    }}int main() {    for (int i = 1; i < N; i++)        sum2[i] = sum2[i - 1] + i;    while (~scanf("%d%d%I64d", &n, &m, &q)) {        for (int i = 1; i <= n; i++)            h[i].read(i);        sort(h + 1, h + n + 1, cmp);        int tmp = 1;        a[h[1].id] = 1;        for (int i = 1; i <= n; i++) {            if (h[i].v != h[i - 1].v) tmp++;            a[h[i].id] = tmp;        }        for (int i = 0; i < m; i++) scanf("%d%d", &X[i], &Y[i]);        memset(ans, 0, sizeof(ans));        gao(q, 1); gao(q - 1, -1);        for (int i = 0; i < m; i++)            printf("%I64d\n", ans[i]);    }    return 0;}

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