Special-half

Binary search

Binary search :( board left and right to open and close)
for the other half of the case can refer https://blog.csdn.net/CCSGTC/article/details/80586181

In the binary process, always maintain left-right open and closed conditions

void solve(){
    sort(heig,heig+n);
    int left = 0, right = 1000000001; // [0,1000000001)
    while(left+1<right){
        int mid = (left+right)/2;
        if(fuc(mid)) left = mid;
        else right = mid;
    }
    printf("%d\n",left);
}

The final boundary conditions:
(Source) https://oi-wiki.org//basic/binary/

half naked title https://www.luogu.org/problemnew/show/P1873

Half of them the possibility to

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int n,m;
int heig[1000010];

void get_data(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d",&heig[i]);
    }
}

bool fuc(int mid){
    long long ans = 0;
    for(int i=n-1;heig[i]>mid;i--){
        ans += heig[i]-mid;
    }
    return ans>=m;
}

void solve(){
    sort(heig,heig+n);
    int left = 0, right = 1000000001;
    while(left+1<right){
        int mid = (left+right)/2;
        if(fuc(mid)) left = mid;
        else right = mid;
    }

    printf("%d\n",left);
}

int main() {
    get_data();
    solve();
    return 0;
}

Fractional Programming

Basis: https://blog.csdn.net/hzoi_ztx/article/details/54898323

Bare question: https://vjudge.net/problem/POJ-2976

Reminded program exits conditions (all tears ah)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;


int n,k;
int a[1010],b[1010];

void get_data(){
    for(int i=0;i<n;i++) cin>>a[i];
    for(int i=0;i<n;i++) cin>>b[i];
}

bool judge(double mid){
    double result[1010];

    double ans = 0;
    for(int i=0;i<n;i++) result[i] = a[i] - mid*b[i];
    sort(result,result+n);
    for(int i=n-1;i>=k;i--) ans += result[i];
    return ans>=0;
}

void solve(){
    double left = 0, right = 1e8;
    while(right-left>1e-6){
        double mid = (right+left)/2;
        if(judge(mid)) left = mid;
        else right = mid;
    }

    printf("%.0lf\n",100*left);
}


int main(){
    while(cin>>n>>k&&n!=0){  //k是可以等于0而不退出的，，我靠（卡了我2个小时）
        get_data();
        solve();
    }
    return 0;
}