笔试强训(三十二)

一.素数回文

https://www.nowcoder.com/practice/d638855898fb4d22bc0ae9314fed956f?tpId=290&tqId=39945&ru=/exam/oj

#include <iostream>
#include <algorithm>
using namespace std;

void check(long long ans)
{
    if(ans < 2)
    {
        cout << "noprime" << endl;
        return;
    }
    for(long long i = 2;i <= ans / i;i++)
    {
        if(ans % i == 0)
        {
            cout << "noprime" << endl;
            return;
        }
    }
    cout << "prime" << endl;
}

int main() {
    string s;
    cin >> s;
    string ss = s;
    reverse(s.begin(),s.end());
    ss += string(s.begin() + 1,s.end());
    long long ans = atol(ss.c_str());
    check(ans);
    return 0;
}

二.活动安排

https://www.nowcoder.com/practice/16d971e9e42e4f3b9b1e2b8794796a43?tpId=308&tqId=40488&ru=/exam/oj

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 2e5 + 10;
struct node
{
    int begin,end;
}a[N];

bool cmp(const node& x,const node& y)
{
    if(x.end != y.end)
    {
        return x.end < y.end;
    }
    return x.begin < y.begin;
}

int main() {
    int n;
    cin >> n;
    for(int i = 0;i < n;i++)
    {
        cin >> a[i].begin >> a[i].end;
    }
    sort(a,a + n,cmp);
    int sum = 1;
    int end = a[0].end;
    for(int i = 1;i < n;i++)
    {
        if(a[i].begin >= end)
        {
            end = a[i].end;
            sum++;
        }
    }
    cout << sum << endl;
    return 0;
}

三.合唱团

https://www.nowcoder.com/practice/661c49118ca241909add3a11c96408c8?tpId=122&tqId=33652&ru=/exam/oj

#include <iostream>
using namespace std;

typedef long long LL;

const int N = 55,M = 15;
LL a[N];
int k,d;

LL INF = 0x3f3f3f3f3f3f3f;
LL f[N][M],g[N][M];



int main() {
    int n;
    cin >> n;
    for(int i = 1;i <= n;i++)
    {
        cin >> a[i];
    }
    cin >> k >> d;
    for(int i = 1;i <= n;i++)
    {
        g[i][1] = f[i][1] = a[i];
        for(int j = 2;j <= min(i,k);j++)
        {
            f[i][j] = -INF;
            g[i][j] = INF;
            for(int prev = max(i - d,j - 1);prev <= i - 1;prev++)
            {
                f[i][j] = max(max(f[prev][j - 1] * a[i],g[prev][j - 1] * a[i]),f[i][j]);
                g[i][j] = min(min(f[prev][j - 1] * a[i],g[prev][j - 1] * a[i]),g[i][j]);
            }
        }
    }
    LL ret = -INF;
    for(int i = k;i <= n;i++)
    {
        ret = max(ret,f[i][k]);
    }
    cout << ret << endl;
    return 0;
}