二分查找，三分查找

题意：   先输入一个整数，后在输入n个整数（保证）输入的序列为有序序列（递增或递减），最后输入一个整数m,问该序列是否存在m,若存在则输出YES，否则输出NO

简言之就是在一个有序数组里面查找是否存在m

 二分如下  ：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x));
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double eps=1e-12;
int m,a[1010],flag=0;
void binsearch1(int a[],int start,int end) //递归
{
int mid=(start+end)/2;
if (a[mid]==m){
flag=1;
printf("YES ");
}
else if (m>a[mid])
binsearch1(a,mid+1,end);
else if (m<a[mid])
binsearch1(a,start,mid-1);
}
void binsearch2(int a[],int start,int end)
{
while (start<=end&&!flag){
int mid=(start+end)/2;
if (a[mid]>m)
end=mid-1;
else if(a[mid]<m)
start=mid+1;
else{
printf("YES ");
flag=1;
}
}
}
int main()
{
int n;
cin>>n;
for (int i=0;i<n;i++)
cin>>a[i];
cin>>m;
binsearch1(a,0,n-1); //递归版
binsearch2(a,0,n-1); //while循环版
if (flag==0)
cout<<"NO";
return 0;
}

二分查找有有点也有缺陷，优点：大大提高了查找的效率，  缺陷：只能查钊单调递增的序列

假若要查找一个抛物线的最值的时候，这里是二分法是不适用的，这里适用三分。

 HDU 2899

Problem Description
Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
 

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
 

Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
 

Sample Input
2
100
200
 

Sample Output
-74.4291
-178.8534
 

Author
Redow

　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Swap(a,b,t) t=a,a=b,b=t
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double eps=1e-6;
//  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
int y;
double solve(double x)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
int main()
{
    int t;
    cin>>t;
    while (t--){
        scanf("%d.%s",&num,&str);
        cin>>y;
        double l=0.0,r=100.0;
        while (r-l>=eps){
            double mid=(l+r)/2;
            double mmid=(mid+r)/2;        
            if(solve(mid)<solve(mmid)){
                r=mmid;
            }
            else
                l=mid;
        }
        printf("%.4lf
",solve(l));
    }
    return 0;
}