001__c @ 2019-05-31 21:05:46

}

by 001__c @ 2019-05-31 21:08:57

include<bits/stdc++.h>

using namespace std;

int main()

{

int n,q,a[1001],b[1001],k[1001],i,j,l,w[1001],flag;
cin>>n>>q;
for(i=1;i<=n;i++)
cin>>a[i];
sort(a,a+n+1);
for(i=1;i<=q;i++)
cin>>k[i]>>b[i];
for(i=1;i<=q;i++)
w[i]=1;
for(i=1;i<=q;i++)
for(j=1;j<=k[i];j++)
    w[i]=w[i]*10;
for(i=1;i<=q;i++)
{       
    for(l=1;l<=n;l++)
    {
        if(a[l]%w[i]==b[i])
        {
            cout<<a[l]<<endl;
            break;
        }
        else if(l==n)
        cout<<"-1"<<endl;
    }   

}

return 0; }

by foxdemon @ 2019-06-01 11:42:18

#include<bits/stdc++.h>
using namespace std;

int n ;//书数量 
int q ;//读者数量 
int b[10005] ;//书码 
int r[10005] ;//需求码 
int l[10005] ;//长度 
long long s  ;
bool flag    ;

int main()
{

    cin >> n >> q ;
    for ( int i = 1 ; i <= n ; i++ )
    {
        cin >> b[i] ;
    }
    sort ( b + 1 , b + n + 1 ) ;
    for ( int i = 1 ; i <= q ; i++ )
    {
        flag = 0 ;
        cin >> l[i] >> r[i] ;
        long long s = pow ( 10 , l[i] ) ;
        for ( int j = 1 ; j <= n ; j++ )
        {
            if ( b[j] % s  == r[i] )
            {
                cout << b[j] << endl ;
                flag = 1 ;
                break ;
            }
        }
        if ( flag == 0 )
        {
            cout << "-1" << endl ;
        }
    }

    return 0;
}

by foxdemon @ 2019-06-01 11:43:05

字符串解法

#include<bits/stdc++.h>
using namespace std ;

string s[1005] ;
string x ;
int n , q , l ;
bool flag , flag1 ;
bool com( string a , string b )
{
    int len1 = a.size() ;
    int len2 = b.size() ;
    if ( len1 == len2 ) return a < b ;
    return len1 < len2 ;
}

int  main()
{
    cin >> n >> q ;
    for ( int i = 1 ; i <= n ; i++ )
    {
        cin >> s[i] ;
    }
    sort ( s + 1 , s + 1 + n , com ) ;
    for ( int i = 1 ; i <= q ; i++ )
    {
        cin >> l >> x ;
        flag1 = 0 ;
        for ( int j = 1 ; j <= n ; j++ )
        {
            flag = 0 ;
            int len = s[j].size() - 1 ;
            for ( int k = l - 1 ; k >= 0 ; k-- , len-- )
            {
                if ( x[k] != s[j][len] )//不成立 
                {
                    flag = 1 ;
                    break ;
                }
            }
            if ( flag == 0 )//如果成立 
            {
                flag1 = 1 ;
                cout << s[j] << endl ;//输出 
                break ;
            }
        }
        if ( flag1 == 0 )//如果没有 
        {
            cout << -1 << endl ;
        }

    }

    return 0 ;

}

by foxdemon @ 2019-06-01 11:43:41

@001__c 您康康,应该是差不多的,应该找的到bug辽QWQ