F. The Number of Products

F. The Number of Products

https://codeforces.com/group/5yyKg9gx7m/contest/274333/problem/F

You are given a sequence a1,a2,…,an consisting of n integers.

You have to calculate three 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 zero; the number of pairs of indices (l,r) (l≤r) such that al⋅al+1…ar−1⋅ar is positive; Input The first line contains one integer n (1≤n≤2⋅105) — the number of elements in the sequence.

The second line contains n integers a1,a2,…,an (−109≤ai≤109) — the elements of the sequence.

Output Print three integers — the number of subsegments with negative product, the number of subsegments with product equal to zero and the number of subsegments with positive product, respectively.

Examples input 5 5 -3 3 -1 0 output 6 5 4 input 10 4 0 -4 3 1 2 -4 3 0 3 output 12 32 11 input 5 -1 -2 -3 -4 -5 output 9 0 6 Note In the first example there are six subsegments having negative products: (1,2), (1,3), (2,2), (2,3), (3,4), (4,4), five subsegments having products equal to zero: (1,5), (2,5), (3,5), (4,5), (5,5), and four subsegments having positive products: (1,1), (1,4), (2,4), (3,3).

分析：

分3个数组，pos[i]表示以i结尾的区间积为正数的数量，nage[i]表示负数，zero[i]表示区间积为0。

1.如果第i为是正数:

pos[i]=pos[i-1]+1。加一是因为包含自己一个数。

nage[i]=nage[i-1]。

zero[i]=zero[i-1]。

2.如果第i为是负数：

nage[i]=pos[i-1]+1。负数可由正数乘上第i变成负数。加一是加上自己。

pos[i]=nage[i-1]。负数可由正数乘上第i变成正数数。

zero[i]=zero[i-1]。

3.如果第i是0：

pos[i]=0; nage[i]=0; 清0，因为乘上0都变成0了。 zero[i]=zero[i-1]+pos[i-1]+nage[i-1]+1; 正数或负数或者0乘上0都是0。

最后再求一下和。

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<sstream>
#include<vector>
#include<stack>
#include<deque>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=2e5+6;
const int mod=1e4+7;
int pos[maxn],nage[maxn],o[maxn];
int main()
{
    int n;
    cin>>n;
    //以i为定点，前面的区间数 
    for(int i=1;i<=n;i++)
    {
        int a;
        scanf("%d",&a);
        if(a>0)
        {
            pos[i]=pos[i-1]+1;
            nage[i]=nage[i-1];
            o[i]=o[i-1];
        }
        else if(a<0)
        {
            pos[i]=nage[i-1];
            nage[i]=pos[i-1]+1;
            o[i]=o[i-1];
        }
        else 
        {
            pos[i]=0;
            nage[i]=0;
            o[i]=o[i-1]+pos[i-1]+nage[i-1]+1;
        }
    }
    ll poss=0,nag=0,zeo=0;
    for(int i=1;i<=n;i++)
    {
        poss+=pos[i];
        nag+=nage[i];
        zeo+=o[i];
    } 
    printf("%lld %lld %lld\n",nag,zeo,poss);
    return 0;
}

F. The Number of Products

原文：https://www.cnblogs.com/studyshare777/p/12615324.html