0分求解

lihaoran2012 @ 2024-08-18 14:04:00

#include<bits/stdc++.h>
using namespace std;
const long long N=1e8+10;
long long d[N],b[N]; 
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    cin>>d[i];
    for(int j=1;j<=m;j++)
    cin>>b[j];
    for(int i=1;i<=m;i++){
        int cnt=0,ans=0;
        for(int j=1;j<=n;j++){
            ans++;
            if(b[i]==d[j]){
                ++cnt;
            }
            if(ans==m){
                break;
            }

        }
        if(cnt==0){
            cout<<"-1"<<" ";
        }else{
            cout<<cnt<<" ";
        }
    }
return 0;
}

by Emil_ @ 2024-08-18 14:06:17

@lihaoran2012 ac代码：

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int N=1e6+100;
int n,m,a[N],p;
int main(){
    std::ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }

    for(int i=1;i<=m;i++){
        cin>>p;
        int ans=lower_bound(a+1,a+1+n,p)-a;
        if(p!=a[ans])
            cout<<-1<<" ";
        else
            cout<<ans<<" ";
    }
    return 0;
}

求关

by 违规用户名K&xs3Z^ @ 2024-08-18 14:06:18

@lihaoran2012 读入数据较多且不小 使用二分和scanf加快速度

#include<bits/stdc++.h>
using namespace std;
int n,a[1114514],l,r,sum,k;
int fun(int l,int r,int sum){
    while(l<=r){
    int mid=r+l>>1;
    if(a[mid]==sum) r=mid-1;
    else if(a[mid]<sum) l=mid+1;
    else if(a[mid]>sum) r=mid-1;
    }
    if(a[l]==sum) return l;
    else return -1;
}
int main(){
    scanf("%d",&n);
    scanf("%d",&k);
    for(int i=1;i<=n;i++){
    scanf("%d",&a[i]);
    }
    for(int i=1;i<=k;i++){
        scanf("%d",&sum);
        cout<<fun(1,n,sum)<<" ";
    }
    return 0;
}

by lihaoran2012 @ 2024-08-18 14:18:12

@Emil_ 已关注。