CodeForces 1474D :Cleaning 思维 + 前缀和

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传送门

题意

分析

分析起来有点前缀和的味道？
假设我们要去交换$i$和$i+1$，那么，我们考虑$1-i-1$区间最后剩下的数$x$和$i+2-n$区间最后剩下的数$y$，只要能够保证$a[i+1]-x==a[i]-y$即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int a[N],b[N],c[N];
int n;int main() {int T;read(T);while(T--){read(n);for(int i = 1;i <= n;i++) read(a[i]),b[i] = c[i] = 0;a[n + 1] = b[n + 1] = c[n + 1] = 0;for(int i = 1;i <= n;i++){if(a[i] < b[i - 1] || b[i - 1] == -1) b[i] = -1;else b[i] = a[i] - b[i - 1];} for(int i = n;i;i--){if(a[i] < c[i + 1] || c[i + 1] == -1) c[i] = -1;else c[i] = a[i] - c[i + 1];}if(!b[n]){puts("YES");continue;}bool flag = false;for(int i = 1;i < n;i++){if(b[i - 1] == -1 || c[i + 2] == -1) continue;if(a[i + 1] < b[i - 1] || a[i] < c[i + 2]) continue;if(a[i] - c[i + 2] == a[i + 1] - b[i - 1]) flag = true;}if(flag) puts("YES");else puts("NO");}}

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