思路：二分之后用dp去check就好啦。

#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<LL, int> #define ull unsigned long long using namespace std;const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8;int n, a[N], mx[N], mn[N];bool check(int x) {if(a[1] > x) return false;mx[1] = mn[1] = a[1];for(int i = 2; i < n; i++) {if(x - a[i-1] < a[i]) return false;mx[i] = min(a[1]-mn[i-1], a[i]);if(x-a[i-1]-a[1]+mx[i-1] >= a[i]) mn[i] = 0;else mn[i] = a[i] - (x-a[i-1]-a[1]+mx[i-1]);}int cnt = a[1] + a[n-1] - mx[n-1];return x - cnt >= a[n]; }int main() {scanf("%d", &n);for(int i = 1; i <= n; i++)scanf("%d", &a[i]);if(n == 1) printf("%d\n", a[1]);else if(n == 2) printf("%d\n", a[1]+a[2]);else {int l = 1, r = 300000, ans = 300000;while(l <= r) {int mid = l + r >> 1;if(check(mid)) r = mid - 1, ans = mid;else l = mid + 1;}printf("%d\n", ans);}return 0; }/* */

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轉載于:https://www.cnblogs.com/CJLHY/p/9840659.html

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