二分搜索

poj1064http://poj.org/problem?id=1064

浮点数最好不要设ub-lb>eps，不好掌握。注意printf(%.2lf",...)是四舍五入，这里不行所以要先取证floor

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define IO ios::sync_with_stdio(false);cin.tie(0);
#define INF 1e9
#define MAXN 10010
const int MOD=1e9+7;
typedef long long ll;
using namespace std;
int n, k; 
double a[MAXN];
int C(double x)
{
    int num = 0;
    for(int i = 0; i < n; i++){
        num += (int)(a[i]/x);
    }
    return num >= k;
}
void solve()
{
    double lb=0, ub=INF;
    for(int i = 0; i < 100; i++){//这里限定次数而不是限定ub-lb>EPS,因为EPS设置不好会陷入死循环 
        double mid = (lb+ub)/2; 
        if(C(mid)){//左闭右开 
            lb = mid;
        } 
        else ub = mid;
    }
    printf("%.2lf
", floor(lb*100)/100);//floor是向下取整，但似乎返回的还是原类型（这里即double），如果直接输出ub则是四舍五入 
}
int main()
{ 
    IO;
    cin >> n >> k;
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    solve();
    return 0;
}

 poj2456http://poj.org/problem?id=2456

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define IO ios::sync_with_stdio(false);cin.tie(0);
#define INF 1e9
#define MAXN 10010
const int MOD=1e9+7;
typedef long long ll;
using namespace std;
int n, m, a[100010];
int C(int x)
{
    int crt=0, last=0;
    for(int i = 1; i < m; i++){//注意这里因为默认第一头奶牛放在第一个牛舍，所以计数从1开始 
        crt=last+1;
        while(crt<n&&a[crt]-a[last]<x){
            crt++;
        }
        last=crt;
        if(crt==n) return 0;
    }
    return 1;
}
void solve()
{
    sort(a, a+n);
    int lb=0, ub=INF;
    while(ub-lb>1){
        int mid = (lb+ub)>>1;
        if(C(mid)){//左闭右开 
            lb = mid;
        }
        else ub = mid;
    }
    cout << lb << endl;
}
int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; i++){
        cin >> a[i];
    } 
    solve();
    return 0;
}