标签：get inpu using syn dash ram cat pen 前缀和
链接：https://codeforces.com/contest/1215/problem/B
You are given a sequence a1,a2,…,ana1,a2,…,an consisting of nn non-zero integers (i.e. ai≠0ai≠0).
You have to calculate two following values:
The first line contains one integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the number of elements in the sequence.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109;ai≠0)(−109≤ai≤109;ai≠0) — the elements of the sequence.
Print two integers — the number of subsegments with negative product and the number of subsegments with positive product, respectively.
题意：求乘积为负数和正数的子段个数（没有0）
题解：正数0，负数1，统计前缀和（就是求前缀和奇偶），再从头扫一遍，维护2个桶前缀和为奇数的个数和前缀和为偶数的个数。
#include <bits/stdc++.h> using namespace std; const int maxn=2e5+5; int n; int a[maxn], sum[maxn], cnt[2]; long long pos, neg; int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n; for(int i=1; i<=n; i++) { cin>>a[i]; if(a[i]>0) a[i]=0; else a[i]=1; } for(int i=1; i<=n; i++) sum[i]=sum[i-1]+a[i]; for(int i=1; i<=n; i++) sum[i]=sum[i]%2; cnt[0]++; //注意cnt0初始化为1，表示区间长度为1的那个 for(int i=1; i<=n; i++) { if(sum[i]){ pos+=cnt[1]; neg+=cnt[0]; } else{ pos+=cnt[0]; neg+=cnt[1]; } cnt[sum[i]]++; } cout<<neg<<" "<<pos; return 0; }
CF 1215 B The Number of Products(思维题)
标签：get inpu using syn dash ram cat pen 前缀和
原文地址：https://www.cnblogs.com/Yokel062/p/11613839.html
