牛客小白月賽98 A~D

牛客小白月賽98 A~D

A-骰子魔術

簽到不多說

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;

int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    int n,x; cin>>n>>x;
    bool ok = false;
    for(int i = 1;i <= n; i++)
    {
        int t; cin>>t;
        if(x==t)ok = true;
    }   
    cout<<(ok?"YES":"NO")<<"\n";

    return 0;
}

B-最少剩幾個？

思路：奇數+偶數=奇數，奇數*奇數=奇數。那麼優先用奇數和偶數配對，剩下的奇數兩兩配對。

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;

int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    int n; cin>>n;
    int even = 0,odd = 0;
    for(int i = 1;i <= n; i++)
    {
        int x; cin>>x;
        even += (x%2==0);
        odd += (x%2!=0);
    }

    int res = min(even,odd);
    odd -= res;
    res += odd/2*2;

    cout<<n-res<<'\n';

    return 0;
}

C-兩個函式

思路：\(\sum_{i = 1}^{x-1}f(f(i)) = \sum_{i = 1}^{x-1}a(ax) = a^2\sum_{i = 1}^{x-1}x = a^2\dfrac{x(x-1)}{2}\)。取模小心點就行了。

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 2e5 + 10;
ll qmi(ll a, ll b, ll mod)
{
    ll ans = 1 % mod;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    int t; cin>>t;
    ll inv = qmi(2, mod - 2, mod);
    while(t--)
    {
        ll a,x;
        cin>>a>>x;
        if(x==1){
            cout<<a*x%mod<<"\n";
            continue;
        }
        ll ans = a*a%mod;
        ans = ans*x%mod;
        ans = ans*(x-1)%mod;
        if(ans<0) ans += mod;
        ans %= mod;
        ans = ans*inv%mod;
        cout<<ans%mod<<"\n";
    }


    return 0;
}

D-切割 01 串 2.0

思路：發現資料範圍很小呀，直接爆搜，加個記憶化。

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 5e3 + 10;
int n,L,R;
string s;
ll s0[N],s1[N],dp[N][N];

ll dfs(ll l,ll r)
{
    ll& res = dp[l][r];
    if(res != -1)
        return res;
    res = 0;
    for(int i = l;i < r; i++)
    {
        ll t = 0;
        ll c0 = s0[i]-s0[l-1];
        ll c1 = s1[r]-s1[i];
        if(abs(c0-c1)>=L && abs(c0-c1)<=R)
            t = 1ll + dfs(l,i)+dfs(i+1,r);
        res = max(res,t);
    }
    return res;
}

int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    cin>>n>>L>>R;
    cin>>s;

    s = "?"+s;
    for(int i = 1;i <= n; i++)
    {
        s0[i] = s0[i-1]+(s[i]=='0');
        s1[i] = s1[i-1]+(s[i]=='1');
    }

    memset(dp,-1,sizeof(dp));
    cout<<dfs(1,n)<<"\n";

    return 0;
}

好像很多人用區間dp。不過也差不多

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 5e3 + 10;
int n,L,R;
string s;
ll s0[N],s1[N],dp[N][N];


int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    cin>>n>>L>>R;
    cin>>s;

    s = "?"+s;
    for(int i = 1;i <= n; i++)
    {
        s0[i] = s0[i-1]+(s[i]=='0');
        s1[i] = s1[i-1]+(s[i]=='1');
    }

    for(int len = 1;len <= n; len++){
        for(int l = 1;l+len-1<=n;l++)
        {
            int r = l+len-1;
            for(int k = l;k < r; k++){
                int t1 = s0[k]-s0[l-1];
                int t2 = s1[r]-s1[k];
                if(abs(t1-t2)>=L && abs(t1-t2)<=R)
                {
                    dp[l][r] = max(dp[l][r],dp[l][k]+dp[k+1][r]+1);
                }
            }
        }
    }
    cout<<dp[1][n];
    return 0;
}