Binary search (example C++)

Binary search

Example: A thinks a number between 1-1000 in his mind, B guesses it, can ask questions, A can only answer yes or no. How to guess the least number of questions?

1. Sequential search: Is it 1? Is it 2?… Is it 999? Ask 500 times on average
2. Binary search: Is it greater than 500? Is it greater than 750? Is it greater than 625?… It only takes 10 times to reduce the guess range to half of the last time
3. The content of binary search must be in order.

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Example 1: Write a function BinarySeach to find the element p in an int array a containing size elements, sorted from small to large, if found, return the element index, if not found, return -1. Requires complexity O(log(n)).

Code:

int BinarySeach(int a[],int size,int p){
    
    
	int L = 0;	//查找区间的左端点
	int R = size - 1;	//查找区间的右端点
	while(L <= R){
    
    	//如果查找区间不为空就继续查找 
		int mid = L+(R-L)/2;	//取查找区间正中元素的下标 
		if(p == a[mid])
			return mid;
		else if(p > a[mid])	//说明p在a[]的左半部 
			L = mid + 1;	//重置查找区间的左端点 
		else 
			R = mid - 1;	// 重置查找区间的右端点
	}
	return -1;
}	

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Example 2: Write a function LowerBound to find the element with the largest subscript that is smaller than the given integer p in the int array a containing size elements and sorted from small to large. If found, return its subscript, if not found, return -1.

Code:

int LowerBound(int a[],int size,int p){
    
    
	int L = 0;	//查找区间的左端点 
	int R = size - 1;	//查找区间的右端点 
	int lastPos = -1;	//目前为止找到的最优解 
	while(L <= R){
    
    	//如果查找区间正中元素的下标
		int mid = L+(R-L)/2;	//去查找区间正中元素的下标
		if (a[mid] >= p)
			R = mid - 1;
		else{
    
    
			lastPos = mid;
			L = mid+1;
		}
	}
	return lastPos;
}

Note:
int mid = (L+R)/2; //取查找区间正中元素的下标
To prevent (L+R) from overflowing too much:
int mid = L+(R-L) /2;

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Example 3: Finding the roots of the equation by dichotomy Find a root of the
following equation: f(x) = x ³ -5x ² +10x-80 = 0; If the root to be found is a, then |f(a)|< = 10 ^-6 .

Solution: Take the derivative of f(x) and get f'(x)=3x ² -10x+10. The formula for finding the roots of a quadratic equation in one variable knows that the equation f'(x) = 0 has no solution, so f'(x) is always greater than 0. Therefore, f(x) is monotonically increasing. It is easy to know that f(0) <0 and f(100)>0, so there must be one and only one root in the interval [0, 100]. Since f(x) is monotonic in [0, 100], we can find the root in the interval [0, 100] by using a bisection method.

Code

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

double EPS = 1e-6;
double f(double x){
    
    
	return x*x*x - 5*x*x + 10*x - 80; 
}
int main(){
    
    
	double root, x1 = 0,x2 = 100,y;
	root = x1+(x2-x1)/2;
	int triedTimes = 1;//记录一共尝试多少次
	y = f(root);

	while( fabs(y) > EPS){
    
    
		if(y > 0)
			x2 = root;
		else 
			x1 = root;
		root = x1+(x2-x1)/2;
		y = f(root);
		triedTimes++;
	}
	printf("%.08f\n",root);
	printf("%d",triedTimes);
	return 0;
}