关于C++中二分法详解

bool check(int x) {/* ... */} // 检查x是否满足某种性质

// 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用：
int bsearch_1(int l, int r)
{
    while (l < r)
    {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;    // check()判断mid是否满足性质
        else l = mid + 1;
    }
    return l;
}
// 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
int bsearch_2(int l, int r)
{
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    return l;
}

#include<iostream>

using namespace std;
const int N = 100000+10;
int q[N];

int main()
{
    int n, m;
    cin>> n >> m;
    
    for(int i = 0; i < n; i++) cin>>q[i];
    while(m--)
    {
      int x;
      cin>> x;
      // 寻找起始位置
      int l = 0, r = n - 1;
      while(l < r)
      {
          int mid =(l + r)/2;
          if(q[mid] >= x) r = mid;
          else l = mid + 1;
      }
      if(q[l] != x) cout<<"-1 -1"<<endl;
      else{
          cout<<l<<" ";
          // 寻找终点位置
          int l = 0, r = n - 1;
          while(l<r)
          {
              int mid = (l + r + 1)/2;
              if(q[mid] <= x) l = mid;
              else r = mid - 1;
          }
          
          cout<< l << endl;
      }
       
    }
    return 0;
}

class Solution {
public:
    int getMissingNumber(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        int l = 0, r = nums.size() - 1;
        while(l < r)
        {
            int mid = (l + r)/2;
            if(nums[mid] != mid) r = mid; //在红色半边(满足条件)
            else l = mid + 1;
        }
        
        //缺的是n这个数
        if(nums[r] == r) r++;
        
        return r;
        
    }
};

浮点数二分算法模板 —— 模板题 AcWing 790. 数的三次方根
bool check(double x) {/* ... */} // 检查x是否满足某种性质（包含了计算和条件）

double bsearch_3(double l, double r)
{
    const double eps = 1e-6;   // eps 表示精度，取决于题目对精度的要求(一般比题目要求的大2)
    while (r - l > eps)
    {
        double mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }
    return l;
}

#include<iostream>

using namespace std;

int main()
{   
    double n;
    cin>>n;
    
    double l = -10000, r = 10000;
     // eps 表示精度，取决于题目对精度的要求(保险1e-8)
    const double eps = 1e-8;  
    while(r - l > eps)
    {
        double mid = (l + r) / 2;
        if(mid * mid * mid >= n) r = mid;
        else l = mid;
    }
    
     printf("%.6lf\n", l);
    return 0;
    
}

#include<iostream>
#include<cstdio>

using namespace std;

double a, b, c, d;// 全局变量方便在cal中使用
const double eps = 1e-6;// 定义精度

//计算一元三次方程
double cal(double x)
{
    return a*x*x*x + b*x*x + c*x + d;
}
int main()
{
    cin>>a>>b>>c>>d;
    
    //枚举根
    for(int i = -100; i <= 100; i++)
    {
        //根与根之差的绝对值 ≥1
        double l = double(i), r = double(i + 1);// 细节:要将l,r转为double
        if(cal(l) == 0) printf("%.2lf ", l); //若f(x) = 0，根即为x 
        
        //f(x1)×f(x2) < 0 根在（x1,x2）之间—— 浮点二分
        else if(cal(l) * cal(r) < 0)
        {
            while(r - l > eps)
            {
                //x1 < x，f(x1)×f(x2)<0，则在(x1, x2)之间一定有一个根
                double mid = (l + r)/2;
                // check()条件
                if(cal(l) * cal(mid) <= 0) r = mid;
                else l = mid;
            }
            
            printf("%.2lf ", l);
        }
    }
}

#include <bits/stdc++.h>
using namespace std;

double check(double x)
{
    return 7*x*x*x*x + 5*x*x*x + 11*x + 6;
} 
double erfen(double Y)
{
    double l=0.0, r=99.0, mid;
    while(r - l > 1e-6){
        mid = (l + r)/2;
        if(check(mid) > Y) r = mid;
        else l = mid;
    }
    return mid;
}
int main()
{
    double Y;
    while(~scanf("%lf", &Y)){
        if(Y < 6 || Y > 677269824)
            puts("None");
        else
        printf("%.4f\n", erfen(Y));
    }
    return 0;
}