codeforces1474D. Cleaning

https://codeforces.com/contest/1474/problem/D

先考虑不能变的情况，由于a[1]要被删光，那么一定是a[1]-=a[1],a[2]-=a[1],依次类推b[i]=a[i]-b[i-1]，能消除的充要条件是所有b[i]>=0且b[n]=0

然后我们考虑枚举每一个相邻位置看可不可行，我们发现相邻位置变化对于奇数i的b[i]和偶数i的b[i]的改变是一样的

那么就只要看一下1-i-1这段b[i]是不是都大于0，再看看i-n这段区间的奇数小标的最小值和偶数下标的最小值变化后是否是>=0的，然后再看看b[n]的变化是否会变成0就可以了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;const int maxl=3e5+10;int n,m,k,cnt,tot,cas,ans;
ll a[maxl],b[maxl];
ll mi[21][2][maxl];
bool vis[maxl];
char s[maxl];inline void prework()
{scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%lld",&a[i]);b[i]=a[i]-b[i-1];}for(int i=1;i<=n;i++)mi[0][i&1][i]=mi[0][i&1][i-1]=mi[0][i&1][i+1]=b[i];for(int k=1;k<=20;k++)for(int d=0;d<=1;d++)for(int i=1;i<=n;i++)mi[k][d][i]=min(mi[k-1][d][i],mi[k-1][d][i+(1<<(k-1))]);
}inline ll getmi(int l,int r,int d)
{int k=log2(r-l+1);return min(mi[k][d][l],mi[k][d][r-(1<<k)+1]);
}inline void mainwork()
{ans=0;if(getmi(1,n,0)>=0 && getmi(1,n,1)>=0 && b[n]==0){ans=1;return;}for(int i=1;i<=n-1;i++){bool flag=true;if(i-1>0 && getmi(1,i-1,(i-1)&1)<0)flag=false;if(i-2>0 && getmi(1,i-2,(i-2)&1)<0)flag=false;if(b[i]-a[i]+a[i+1]<0)flag=false;if(i+2<=n && getmi(i+2,n,i&1)+2*a[i+1]-2*a[i]<0)flag=false;if(getmi(i+1,n,(i+1)&1)+2*a[i]-2*a[i+1]<0)flag=false;if((i&1)==(n&1) && b[n]+2*a[i+1]-2*a[i]!=0)flag=false;if((i&1)!=(n&1) && b[n]+2*a[i]-2*a[i+1]!=0)flag=false;if(flag){ans=1;return;}}
}inline void print()
{puts(ans?"YES":"NO");
}int main()
{int t=1;scanf("%d",&t);for(cas=1;cas<=t;cas++){prework();mainwork();print();}return 0;
}

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