题目描述

本题大致意思就是求一个数组有哪些区间含奇数个负数

算法1

DP Solution

DP[i] 表示以 i 结尾的区间有多少个含奇数个负数
- 则若 xs[i] < 0，则接到 i - 1 含偶数个负数的区间上
- 否则，则接到 i - 1 含奇数个负数的区间上

时间复杂度

O(n)

C++ 代码

#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <tuple>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;

template <class T>
T read() {
    T x;
    cin >> x;
    return x;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n = read<int>();
    vector<int> xs;
    for (int i = 0; i < n; i += 1) { xs.push_back(read<int>()); }

    vector<long> dp(xs.size());
    dp[0] = xs[0] < 0;
    for (int i = 1; i < xs.size(); i += 1) {
        dp[i] = xs[i] < 0;

        if (xs[i] < 0) {
            dp[i] += i - dp[i - 1];
        } else {
            dp[i] = dp[i - 1];
        }
    }

    long ncnt = 0, pcnt = 0;
    for (int i = 0; i < xs.size(); i += 1) {
        ncnt += dp[i];
        pcnt += i + 1 - dp[i];
    }
    cout << ncnt << " " << pcnt << endl;

    return 0;
}