題干：

給定N朵花的原先的高度，從左到右排列，最多澆水m天，每天只能澆一次，每次使得連續的w朵花的高度增長1，問最后最矮的花的高度最高是多少。

Examples

Input

6 2 3 2 2 2 2 1 1

Output

2

Input

2 5 1 5 8

Output

9

Note

In the first sample beaver can water the last 3 flowers at the first day. On the next day he may not to water flowers at all. In the end he will get the following heights: [2, 2, 2, 3, 2, 2]. The smallest flower has height equal to 2. It's impossible to get height 3 in this test.

解題報告：

? 直接二分即可。樹狀數組差分維護區間更新，復雜度O(nlognlogn)，其實可以優化到nlogn，直接用一個變量維護增量即可。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 4e5 + 5; ll n,m,w,a[MAX]; ll c[MAX]; int lowbit(int x) {return x&-x;} ll sum(int x) {ll res = 0;while(x) {res += c[x];x -= lowbit(x);} return res; } void update(int x,ll val) {while(x < MAX) {c[x] += val;x += lowbit(x);} } bool ok(ll x) {ll cnt = 0,tmp;for(int i = 1; i<=n+w+1; i++) c[i]=0;for(int i = 1; i<=n; i++) {tmp = sum(i);if(a[i] + tmp < x) {cnt += (x-a[i]-tmp);update(i,x-a[i]-tmp);update(i+w,-(x-a[i]-tmp));}} return cnt <= m; } int main() {cin>>n>>m>>w; for(int i = 1; i<=n; i++) scanf("%lld",a+i);ll l = 0,r = 2e9,mid,ans;while(l<=r) {mid = (l+r)>>1;if(ok(mid)) l = mid+1,ans = mid;else r = mid-1;}printf("%lld\n",ans);return 0 ; }

?

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