Let 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]
is converted to all prefixes and minimum values>0
1️⃣.k>=n/2 ----> s[1]=a[1]+.,…+ a[k], s[2]=xa[1], s[3]=xa[2];
take s[1]->0 after subtracting the prefix min M[ ], then a[1] +…+a[k]+M[n-k+1] is the minimum value
2️⃣k<n/2
if s[i]=a[i]+…+a[i+k-1]>0,s [i+k]=a[i+k]+…+a[i+k+k-1]>0
Then s[i+k*2]>0 can be summarized in 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");
}