POJ-2018 Best Cow Fences 二分

题意：找到一个连续区间，区间的长度至少大于f，现在要求这个区间的平均值最大。

题解：

二分找答案。

每次对于2分的mid值， 都把原来的区间减去mid， 然后找到一长度至少为f的区间， 他们的区间和>=0。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<map>
 #include<iostream>
 #include<cstring>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = 1e5 + ;
 int n, f;
 int a[N];
 LL suf[N];
 bool check(int mid){
     for(int i = ; i <= n; ++i)
         suf[i] = suf[i-] + a[i] - mid;
     LL minval = ;
     for(int i = f; i <= n; ++i){
         minval = min(minval, suf[i-f]);
         if(suf[i] - minval >= ) return true;
     }
     return false;
 }
 int main(){
     scanf("%d%d", &n, &f);
     for(int i = ; i <= n; ++i)
         scanf("%d", &a[i]), a[i] *= ;
     int l = , r = inf;
     while(l <= r){
         int m = l+r >> ;
         if(check(m)) l = m + ;
         else r = m - ;
     }
     printf("%d\n", l-);
     return ;
 }