题目描述

超时了，怎么办

样例

#include<iostream>
using namespace std;
const int N = 100010;
int q[N];
int n,m;
int lower_bound(int q[],int target){
    int l = -1, r = n,mid;
    while(l+1<r){
        mid = (r+l) >> 1 + l;
        if(q[mid]<target){
            l = mid;
        }else{
            r = mid;
        }
    }
    return r;

}
int main(){
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ ){
        scanf("%d", &q[i]);
    }
    while(m--){
        int target;
        cin>>target;
        int a = lower_bound(q,target);
        if (a>=n || q[a]!=target){
            cout<<"-1"<<" "<<"-1"<<endl;
            continue;
        }
        int b = lower_bound(q,target+1)-1;
        cout<<a<<" "<<b<<endl;
    }
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla