牛客小白月赛88

超级闪光牛可乐

直接暴力即可

// by Totoro
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
const int N = 1e6 + 100;
int main()
{
    int x, n;
    cin >> x >> n;
    int maxn=0;
    char y;
    rep(i, 1, n)
    {
        char c;
        cin >> c;
        ll a;
        cin >> a;
        if(a>maxn)
        {
            maxn=a;
            y=c;
        }
    }
if(x/maxn>1000)
{
    cout<<-1;
}
else
{
    if(x%maxn)
    {
x+=maxn;
    }
    rep(i,1,x/maxn)
    {
        cout<<y;
    }
}
return 0;
}

人列计算机

// by Totoro
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
const int N = 1e6 + 100;
void solve()
{
        vector<string> s(5);
        for (int i = 0; i < 5; i++)
        {
            getline(cin, s[i]);
        }

        char op = s[2][5];
        if (op == '&')
        {
            cout << (int(s[1][0]-'0') & int(s[3][0] - '0'))<< "\n";
        }
        else if (op == '=')
        {
            cout << (int(s[1][0]-'0') | int(s[3][0] - '0')) << "\n";
        }
        else
        {
            cout << !(s[2][0] - '0') << "\n";
        }
    }
int main()
{
    int t = 1;
    while (t--)
    {
        solve();
    }
}

时间管理大师

// by Totoro
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
const int N = 1e6 + 100;
bool a[N];
int main()
{
int n;
cin>>n;
int ans=0;
rep(i,1,n)
{
    int x,y;
    cin>>x>>y;
    int bell=x*60+y;
    int z,b,c;
    z=bell-1;
    b=bell-3;
    c=bell-5;
    if(!a[z])
    {
ans++;
a[z]=1;
    }
     if(!a[b])
    {
ans++;
a[b]=1;
    }
    if(!a[c])
    {
        ans++;
        a[c]=1;
    }
}
cout<<ans<<'\n';
rep(i,0,24*60)
{
    if(a[i])
    {
        cout<<i/60<<' '<<i%60<<'\n';
    }
}


}

我不是大富翁

// by Totoro
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
const int N = 6000;
bool dp[N][N];
int main()
{
int n,m;
cin>>n>>m;
//dp[i][j]表示第i步能否在J位置
//dp[i][j]=dp[i-1][(j-a[i]+n)%m]|dp[i-1][(j+a[i])%m]
dp[0][0]=1;
rep(i,1,m)
{
    int x;
    cin>>x;
    x%=n;
for(int j=0;j<n;j++)
{
dp[i][j]=dp[i-1][(j+x)%n]|dp[i-1][(j-x+n)%n];
}
}
if(dp[m][0])
{
    cout<<"YES"<<'\n';
}
else
{
    cout<<"NO"<<'\n';
}


}

多重映射

非常妙，记录了每个数最后变成什么，多体会

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
typedef double db;
#define fi first
#define se second
using i128 = __int128_t;
using piii = pair<ll, pair<ll, ll>>;
const ll inf = 1ll << 60;
const int N = 1e6 + 10;
void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    vector<pii> q(m + 1);
    map<int, int> t;
    for (int i = 1; i <= m; i++)
    {
        cin >> q[i].fi >> q[i].se;
        t[q[i].fi] = q[i].fi;
        t[q[i].se] = q[i].se;
    }
    for (int i = m; i > 0; i--)
    {
        t[q[i].fi] = t[q[i].se];
    }
    for (int i = 1; i <= n; i++)
    {
        if (t.find(a[i]) == t.end())
        {
            cout << a[i] << ' ';
        }
        else
        {
            cout << t[a[i]] << ' ';
        }
    }
    cout << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}