Codeforces Round #585 (Div. 2) B. The Number of Products(状态机)
https://codeforces.com/contest/1215/problem/B
给你一个序列a1,a2,…,an,由n个非零整数组成(即ai≠0)。
您必须计算以下两个值:
使得al⋅al+1…ar−1⋅ar为负的指数对(l,r) (l≤r)的个数;
使得al⋅al+1…ar−1⋅ar为正的指数对(l,r) (l≤r)的个数;
输出
打印两个整数—分别是负乘积的子段数和正乘积的子段数。
input
5
5 -3 3 -1 1
output
8 7
input
10
4 2 -4 3 1 2 -4 3 2 3
output
28 27
input
5
-1 -2 -3 -4 -5
output
9 6
状压dp写法,可惜一直Compilation error
实在是不理解wa?
此写法仅供参考
//#include<bits/stdc++.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=500200;
const int M=5002;
const int mod=998244353;
LL a[N],b[N];
//int h[N],w[N],e[N],ne[N],dist[N],idx;
LL f[N][2],dp[N][M];
bool vis[N],st[N];
int dx[]={-1,0,0,1,-1,-1,1,1},dy[]={0,1,-1,0,1,-1,-1,1};
LL minn=0,maxn=0;
int n,d,e;
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
LL n;
cin>>n;
for(LL i=1;i<=n;i++)
cin>>a[i];
LL fu=0,zh=0;
for(LL i=1;i<=n;i++)
{
if(a[i]>0)
{
f[i][1]+=f[i-1][1]+1;//正数个数+1
f[i][0]+=f[i-1][0];//当前负数个数不变
}
else
{
f[i][0]+=f[i-1][1]+1;//负数个数是在前面正数基础上加1
f[i][1]+=f[i-1][0];//正数是在之前的负数的基础上负负得正来的
}
fu+=f[i][0];
zh+=f[i][1];
}
/*for(int i=1;i<=n;i++)
cout<<f[i][0]<<" ";
cout<<endl;
for(int i=1;i<=n;i++)
cout<<f[i][1]<<" ";
cout<<endl;*/
cout<<fu<<" "<<zh<<endl;
}
return 0;
}
附上简单正解
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
/*#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<deque>
typedef pair<int,int> PII;
const int N=200200;
const int M=5002;
const int mod=998244353;
int a[N],b[N];
//int h[N],w[N],e[N],ne[N],dist[N],idx;
int f[N][3],dp[N][M];
bool vis[N],st[N];
int dx[]={-1,0,0,1,-1,-1,1,1},dy[]={0,1,-1,0,1,-1,-1,1};*/
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
LL n;
cin>>n;
LL fu=0,zh=0;
LL num=0,sum=0;
for(int i=1;i<=n;i++)
{
LL x;
cin>>x;
num++;
if(x<0) swap(num,sum);
zh+=num;
fu+=sum;
}
cout<<fu<<" "<<zh<<endl;
}
return 0;
}