满屏RE

Use_Imagination @ 2023-07-02 17:37:35

#include<bits/stdc++.h>
using namespace  std;
int a[100000];
int finnd(int qWq,int x){
    for(int i=0;i<=x;i++){
        if(a[i]==qWq)
            return i;
    }
}
int judge(int k,int qwq){
    int mid=1,pos=0,tail=qwq;
    while(pos<=tail){
        mid=(pos+tail)/2;
        if(a[mid]==k) return a[mid];
        if(a[mid]<k) pos=mid+1;
        if(a[mid]>k) tail=mid-1;
    }
    return -1;
}
int main(){
    int n,m,x;
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=m;i++){
        cin>>x;
        if(judge(x,n)==-1){
            cout<<-1;
            break;
        }
        cout<<finnd(x,n)<<" ";
    }
    return 0;
}

by Gyf117 @ 2023-07-02 17:42:28

int mid=1,pos=0,tail=qwq-1;

by YSR330 @ 2023-07-02 17:43:19

6

by Weizhuo_Zhao @ 2023-07-02 18:10:39

@YSR330 不要无效回复

by Weizhuo_Zhao @ 2023-07-02 18:11:50

看我的AC吧

#include <iostream>
using namespace std;
int nums[1000001];
int main() {
    int n, m, query;
    cin >> n >> m；
    for (int i = 1; i <= n; i++)
        cin >> nums[i];
    for (int i = 0; i < m; i++)
        cin >> query;
        int result = lower_bound(nums+1, nums+n+1, query)-nums;
        cout << ((query == nums[result]) ? result : -1) << " ";
    }
    return 0;
}

by Weizhuo_Zhao @ 2023-07-02 18:12:30

镐的那么复杂

by lorry26 @ 2023-07-02 18:43:00

#include<bits/stdc++.h>
using namespace std;
int hsy,crz,part,first,size,last,sjx;
int main()
{
cin>>hsy>>crz; 
int zzk[hsy+1],lbn[crz+1];
for(int a=1;a<=hsy;a++)
{
cin>>zzk[a];
}
for(int k=1;k<=crz;k++)
{
first=1;
last=hsy;
cin>>sjx;
while(first-last<0)
{
size=last-first;
part=first+size/2;  
if(sjx==zzk[part]) 
{
last=part;
}
else if(zzk[part]<sjx) 
{
first=part+1;
}         
else 
{
last=part-1;
}
}   
if(sjx==zzk[last]) 
{
lbn[k]=last;            
}
else 
{
lbn[k]=-1;  
}  
}
for(int m=1;m<=crz;m++)
{
cout<<lbn[m]<<" ";
}
return 0;
}

这不就好了吗？

by Zzzcr @ 2023-07-02 18:49:05

@lorry26 什么阴间马蜂

by lorry26 @ 2023-07-02 18:55:08

额，本人习惯，没办法

by lorry26 @ 2023-07-02 18:59:54

@Zzzcr