Codeforces - 466C 双指针

Posted 2020-10-17 The Azure Arbitrator

首先要判sum是否是3的整数倍
然后把符合条件的前缀和以及后缀和分别加入到静态vector中
最后扫一下j和k定位在哪然后求总长的差来更新答案
注意i j k至少隔1位,所以lower_bound是s1[i]+2
官方题解是O(n),不过没仔细看
这种解法最坏应该是常数较小的O(nlogn)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5+11;
typedef long long ll;
int a[maxn],n;
ll sumL[maxn],sumR[maxn];
int s1[maxn],s2[maxn];
int main(){
    while(scanf("%d",&n)!=EOF){
        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);
        sumL[0]=sumR[n+1]=0;
        for(int i = 1; i <= n; i++) sumL[i]=sumL[i-1]+a[i];
        for(int i = n; i >= 1; i--) sumR[i]=sumR[i+1]+a[i];
        if(sumL[n]%3!=0){
            printf("0\n");
            continue;
        }
        int cnt1=0,cnt2=0;
        ll ans=0;
        for(int i = 1; i <= n; i++) if(sumL[i]==sumL[n]/3) s1[++cnt1]=i;
        for(int i = 1; i <= n; i++) if(sumR[i]==sumL[n]/3) s2[++cnt2]=i;
        for(int i = 1; i <= cnt1; i++){
            int j = lower_bound(s2+1,s2+cnt2+1,s1[i]+2)-s2;
            int d = cnt2-j+1;
            ans+=d;
        }
        printf("%lld\n",ans);
    }
    return 0;
}