codeforces 645 Are You Fired?

链接

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令s[i] = a[i]+…+a[i+k-1] ,s[i+1] = a[i+k] - a[i], s[i+2] = a[i+k+1] - a[i+1]
则转变为所有前缀和最小值 >0
1️⃣.k>=n/2 ----> s[1]=a[1]+.,…+a[k] , s[2]=x-a[1] , s[3]=x-a[2];
把s[1]->0后取减缀和 的前缀min M[ ],则 a[1]+…+a[k]+M[n-k+1] 即为最小值
2️⃣k<n/2
如果s[i]=a[i]+…+a[i+k-1]>0,s[i+k]=a[i+k]+…+a[i+k+k-1]>0
那么s[i+k*2]>0 则可归纳于1️⃣

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 5;
int n,m;
ll a[maxn],x,pre,b[maxn],c[maxn];
int main() {
    
    
    scanf("%d",&n);
    m=(n+1)/2;

    for(int i=1;i<=m;i++) scanf("%lld",&a[i]);

    scanf("%lld",&x);

    for(int i=m+1;i<=n;i++) a[i]=x;

    for(int i=2;i<=n;i++) c[i]=x-a[i-1]+c[i-1];

    for(int i=2;i<=n;i++) b[i]=min(b[i-1],c[i]);

    m=n/2;
    for(int i=1;i<=m;i++) pre+=a[i];

    for(int k=m+1;k<=n;k++) {
    
    
        pre+=a[k];
        if(pre+b[n-k+1]>0) {
    
    
            printf("%d\n",k);return 0;
        }
    }

    printf("-1\n");
}