P2440 木材加工（二分+贪心）

思路：这里就要看往那边贪心了，因为解决的是最大值最小化，最小值最大化。也就是说当满足大于等于c时，l=mid+1这样的二分得到的就是在所有满足条件函数下的最右端.

 

#include
#include
using namespace std;

#define ll long long
const int maxn = 1e5 + 10;
int a[maxn], n, maxx, ans;
ll c, mid;

bool check(ll x){
    ll sum = 0;
    for (int i = 1; i <= n; ++i)
        sum += a[i] / x;
    return sum >= c;
}

void half(){
    ll l = 1, r = maxx;
    while (l <= r){
        mid = (r + l) >> 1;
        if (check(mid)){ l = mid + 1; }
        else r = mid - 1;
    }
    ans = r;
}

int main(){
    cin >> n >> c;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i]; maxx = max(maxx, a[i]); 
    }
    half();        //二分
    cout << ans << endl;
}

 

转载于:https://www.cnblogs.com/ALINGMAOMAO/p/10453827.html