codeforces 1042d Petya and Array 线段树|树状数组|平衡树

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本文介绍了一种使用splay树解决数组中求和小于特定值的子序列数量问题的方法。通过预处理前缀和并维护splay树，可以高效地计算出所有符合条件的子序列。文章详细阐述了splay树的旋转、插入和查找操作，以及如何利用这些操作解决给定问题。

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D. Petya and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya has an array aa consisting of nn integers. He has learned partial sums recently, and now he can calculate the sum of elements on any segment of the array really fast. The segment is a non-empty sequence of elements standing one next to another in the array.

Now he wonders what is the number of segments in his array with the sum less than tt. Help Petya to calculate this number.

More formally, you are required to calculate the number of pairs l,rl,r (l≤rl≤r) such that al+al+1+⋯+ar−1+ar<tal+al+1+⋯+ar−1+ar<t.

Input

The first line contains two integers nn and tt (1≤n≤200000,|t|≤2⋅10141≤n≤200000,|t|≤2⋅1014).

The second line contains a sequence of integers a1,a2,…,ana1,a2,…,an (|ai|≤109|ai|≤109) — the description of Petya's array. Note that there might be negative, zero and positive elements.

Output

Print the number of segments in Petya's array with the sum of elements less than tt.

Examples

input

Copy

5 4
5 -1 3 4 -1

output

Copy

input

Copy

3 0
-1 2 -3

output

Copy

input

Copy

4 -1
-2 1 -2 3

output

Copy

题目大意：给定数组a和一个数字t，问有多少对i，j使得子序列(i，j)的和小于t。

预处理前缀和sum[i]，对于位置i，即求有多少个j使得sum[i] - sum[j] < t，即sum[i] - t < sum[j]，所以维护下sum数组中有多少个大于sum[i] - t即可，之前都是用线段树写的，这次用splay写了下。

#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;

#define ll long long
#define lson tr[x][0]
#define rson tr[x][1]
const int maxn = 2e5 + 10;
const int INF = 1e9;
int root, tot, n;
ll t, ans;
int  fa[maxn], tr[maxn][2], sz[maxn];
ll sum[maxn], key[maxn];
int num[maxn], a[maxn];

int judge(int x) {
    return tr[fa[x]][1] == x;
}

void pushup(int x) {
    if (x) {
        sz[x] = sz[tr[x][0]] + sz[tr[x][1]] + num[x];
    }
}

void rotate(int x) {
    int y = fa[x], d = judge(x);
    if (tr[y][d] = tr[x][d ^ 1]) fa[tr[y][d]] = y;
    if (fa[x] = fa[y]) tr[fa[y]][judge(y)] = x;
    tr[fa[y] = x][d ^ 1] = y;
    pushup(y);
}

void splay(int x, int k) {
    for (int y; (y = fa[x]) != k; rotate(x))
        if (fa[y] != k) rotate(judge(x) == judge(y) ? y : x);
    pushup(x);
    if (!k) root = x;
}
   
void insert(ll val) {
    int node = root, f = 0;
    while (node && key[node] != val) {
        f = node;
        node = tr[node][key[node] < val];
    }
    if (node) num[node] += 1;
    else {
        node = ++tot;
        if (f) tr[f][key[f] < val] = node;
        fa[node] = f;
        key[node] = val;
        sz[node] = num[node] = 1;
    }
    splay(node, 0);
}

void find(ll x) {
    int r = root;
    while (key[r] != x && tr[r][key[r] < x])
        r = tr[r][key[r] < x];
    splay(r, 0);
}

int main() {
    scanf("%d%I64d", &n, &t);
    tot = root = ans = sum[0] = 0;
    insert(0);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        sum[i] = sum[i - 1] + a[i];
    }
    for (int i = 1; i <= n; i++) {
        find(sum[i] - t);
        if (key[root] > (sum[i] - t)) ans += (sz[tr[root][1]] + num[root]);
        else ans += sz[tr[root][1]];
        insert(sum[i]);
    }
    printf("%I64d\n", ans);
    return 0;
}