牛客练习赛60

(先划划水）
[Toc]
A 题
签到题

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define pb push_back
#define de(x) cerr<<#x<<" = "<<x<<endl
#define __i __int128
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll; 

ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
int n , k, m ;
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}
ll ar[200010];
int f[200010];



int main()
{
        ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
       cin >> n;
       map<ll , ll > mp;
      for(int i=0;i<n;i++) {

          cin >>ar[i];bitset<31 > be(ar[i]);
          for(int i=0;i<=31;i++)
              if(be[i]==1)mp[i]++;

      }
      ll sum =0 ;
      for(int i=0 ;i<n;i++){
          bitset<31>be(ar[i]);
          for(int j=0;j<=31;j++){
              if(be[j])sum +=(1<<j)*mp[j];
          }
          //cout<<num<<endl;
      }
      cout<<sum<<endl;
    return 0;
}

B 题
签到题

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1005;
ll a[maxn],b[maxn];
ll dis[maxn][maxn];
const ll mod = 998244353;
int main()
{
    ll n;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i]>>b[i];
    ll ans = 0;
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++)
        {
            dis[i][j] = abs(a[i] - a[j]) + abs(b[i] - b[j]);
            ans += 1ll*(dis[i][j]*(n - 2));
            ans %= mod;
        }
    }
    cout<<ans<<endl;

    return 0;
}

C 题

题外话

以后记住了这样的递推是DP 了。。

思路

设f[i][j],是前i个字符里面长度为j个的

然后先不考虑重复的那么f[i][j] = f[i-1][j] + f[i-1]j-1
考虑重复的，设置一个判断这个字符是否出现的数组，然后，会发现其实我们只需要减去这个字符上一个位置的f,就行了，（其实就是把上一个相同字符的次数减去，然后换上当前字符的）

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define pb push_back
#define de(x) cerr<<#x<<" = "<<x<<endl
#define __i __int128
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll; 

ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
int n , k, m ;

vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}
ll f[1010][1010];
ll pre[100010];
ll mod = 1e9+7;char s[10010];
int main()
{
        ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
       cin >>n>> m ;
       cin >>s+1;
       f[0][0] =1 ;
       for(int i=1 ;i<=n;i++){
           f[i][0]=1 ;
           for(int j=1 ;j<=i;j++){
                f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
            if (pre[s[i] - 'a'])f[i][j] -= f[pre[s[i] - 'a'] - 1][j - 1];
            f[i][j] %= mod;

           }    pre[s[i]-'a'] = i;
       }
     ll ans =f[n][m];
    if(ans< 0 )ans +=mod ;
    cout<<ans<<endl;
    return 0;
}