二分法查找

二分法适合数据量比较大的适合来用，用来查找数据，所给的数据必须是有序的，可以提前用sort来排好序。

bool erfen(int *a,int left,int right,int num){
	while(left<right){
		int mid=(left+right)/2;
		if(num==a[mid]){
			return true;
		}
		else if(num<a[mid]){
			right=mid;
		}
		else if(num>a[mid]){
			left=mid+1;
		}
	}
	return false;
}

如果找到num，就返回true；如果没有找到num，就返回false。可以根据题目的需求来返回自己所需要的值。

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

A - Linear Search-- Aizu - ALDS1_4_A 

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output

Print C in a line.

Constraints

• n ≤ 10000
• q ≤ 500
• 0 ≤ an element in S ≤ 109
• 0 ≤ an element in T ≤ 109

Sample Input 1

5
1 2 3 4 5
3
3 4 1

Sample Output 1

3

Sample Input 2

3
3 1 2
1
5

Sample Output 2

0

Sample Input 3

5
1 1 2 2 3
2
1 2

Sample Output 3

2

#include<cstdio>
#include<stdio.h>
#include<string>
#include<iostream> 
#include<map>
using namespace std;
#include<algorithm>

bool erfen(int *a,int left,int right,int num){
	while(left<right){
		int mid=(left+right)/2;
		if(a[mid]==num){
			return true;
		}
		else if(a[mid]>num){
			right=mid;
		} 
		else if(a[mid]<num){
			left=mid+1;
		}
	}
	return false;
}

int main(){
	int n,q,a[100005],b[100005],count=0;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	cin>>q;
	for(int i=0;i<q;i++){
		cin>>b[i];
	}
	sort(a,a+n);
	for(int i=0;i<q;i++){
		if(erfen(a,0,n,b[i])){
			count++;
		}
	}
	cout<<count<<endl;
	return 0;
}

一定要记得先排序！！！

下面这两个题都要用到二分法查找。

POJ - 2456 （Aggressive cows ）

HDU - 5248（序列变换）