超详题解

//框架
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll NM=1e5+2; //不多也不少
ll n,m,a[NM];//好习惯
int main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    cin>>m;
    while(m--){
        
    }
    return 0; //好习惯
}

这个相信大家一定会写
不一定

首先定义三变量R and L and pos
后用pos存储第一个>=k的数

ll pos = lower_bound(a+1,a+n+1,k)-a;

重点是二分怎么写
当然暴力也可以AC

二分

while(left<=right) {
    ll mid=left+(right-left)/2;
    if(a[mid]>=k){
        pos=mid;
        right=mid-1;//重点
    }else left=mid+1;//重点
}

pos-1 最小是 1用c1存储
pos 最大是 n 用c2存储

再比较一下
输出结果

就行了

二分代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll NM=1e5+2; //不多也不少
ll n,m,a[NM];
int main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    cin>>m;
    while(m--){
        //重点
        ll k;
        cin>>k;
        ll left=1,right=n;
        ll pos=n+1;//pos第一个>=k的数
        while(left<=right){//lower_bound函数底层逻辑
            ll mid=left+(right-left)/2;
            if(a[mid]>=k){
                pos=mid;
                right=mid-1;
            }else left=mid+1;
        }//其实这样就可以了ll pos = lower_bound(a+1,a+n+1,k)-a;
        ll c1=-1,c2=-1;
        if(pos>0)c1=a[pos-1];
        if(pos<n)c2=a[pos];
        //特判一下
        ll ans = -1;
        if(c1 == -1)ans = c2;
        else if(c2 == -1)ans = c1;
        else{
            ll diff1=abs(c1-k),diff2=abs(c2-k);
            if(diff1<diff2)ans = c1;
            else if(diff1>diff2)ans = c2;
            else ans = min(c1, c2);
        }cout << ans << endl;
    }
    return 0;
}

枚举代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll NM=1e5+5;
ll n,m,a[NM],b[NM];
int main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++)cin >> a[i];
    cin>>m;
    for(int i = 1;i <= m;i++){
        cin>>b[i];
        ll minn = INT_MAX,ans=0;
        for(int j = 1;j <= n;j++){
            if(abs(a[j]-b[i]) < minn){
                ans=a[j];
                minn=abs(a[j]-b[i]);
            }
        }cout<<ans<<endl;
    }
    return 0;
}

虽然能AC但我建议别这样做
数据再大点就得TLE了

真正100%AC超简洁代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll NM=1e5+2;
ll n,m,a[NM];
int main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    cin>>m;
    while(m--){
        ll k;
        cin>>k;
        ll pos=lower_bound(a+1,a+n+1,k)-a,c1=-1,c2=-1,ans = -1;
        if(pos>0)c1=a[pos-1];
        if(pos<n)c2=a[pos];
        if(c1==-1)ans=c2;
        else if(c2==-1)ans=c1;
        else{
            ll diff1=abs(c1-k),diff2=abs(c2-k);
            if(diff1<diff2)ans=c1;
            else if(diff1>diff2)ans=c2;
            else ans=min(c1, c2);
        }cout << ans << endl;
    }return 0;
}