Codeforces Round #585 (Div. 2) B. The Number of Products(DP)

链接:

https://codeforces.com/contest/1215/problem/B

题意:

You are given a sequence a1,a2,…,an consisting of n non-zero integers (i.e. ai≠0).

You have to calculate two following values:

the number of pairs of indices (l,r) (l≤r) such that al⋅al+1…ar−1⋅ar is negative;
the number of pairs of indices (l,r) (l≤r) such that al⋅al+1…ar−1⋅ar is positive;

思路:

DP[i][0]为i位置往前连续的负数, DP[i][1] 为偶数.
最后累计一下.

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5+10;

int a[MAXN];
LL Dp[MAXN][2];
int n;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1;i <= n;i++)
        cin >> a[i];
    if (a[1] > 0)
        Dp[1][1] = 1;
    else
        Dp[1][0] = 1;
    for (int i = 2;i <= n;i++)
    {
        if (a[i] > 0)
        {
            Dp[i][1]++;
            Dp[i][0] += Dp[i-1][0];
            Dp[i][1] += Dp[i-1][1];
        }
        else
        {
            Dp[i][0]++;
            Dp[i][0] += Dp[i-1][1];
            Dp[i][1] += Dp[i-1][0];
        }
    }
    LL sum1 = 0, sum2 = 0;
    for (int i = 1;i <= n;i++)
        sum1 += Dp[i][0], sum2 += Dp[i][1];
    cout << sum1 << ' ' << sum2 << endl;

    return 0;
}