bzoj 1816 二分

思路：二分答案，然后我们贪心地先不填最小的一堆，看在最小的一堆消耗完之前能不能填满其他堆。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg
using namespace std;
 
const int N = 1e5 + ;
const int M = 5e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;
 
int n, m, a[N];
 
bool check(int x) {
    LL sum = ;
    for(int i = ; i <= n; i++)
        if(a[i] < x) sum += x - a[i];
    if(m < sum) return false;
    if(a[] >= x) return true;
    sum -= x - a[];
    if(sum <= a[]) return true;
    return false;
}
 
int main() {
    scanf("%d%d", &n, &m);
    for(int i = ; i <= n; i++)
        scanf("%d", &a[i]);
    sort(a + , a +  + n);
    int l = , r = 1e9, ans = -;
    while(l <= r) {
        int mid = l + r >> ;
        if(check(mid)) l = mid + , ans = mid;
        else r = mid - ;
    }
    printf("%d\n", ans);
    return ;
}