• dp(E. Sleeping Schedule)


    https://codeforces.ml/contest/1324/problem/E

    题意:一天有h(3 <= h <= 2000)个小时,n(1 <= n <= 2000)次睡觉,每次过ai(1 <= ai < h)时间就睡觉,有两种选择在过ai-1时间去睡或过ai时间去睡 , 准备开始睡觉时间在[l , r]( 0 <=l <= r < h)算是好的睡眠。问最多有几个好的睡眠时间。

    解法:二维dp[i][j]表示第i次睡眠,时间为j的最大好睡眠。

    //#include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <stdio.h>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <string.h>
    #include <vector>
    typedef long long ll ;
    #define int ll
    #define mod 1000000007
    #define gcd __gcd
    #define rep(i , j , n) for(int i = j ; i <= n ; i++)
    #define red(i , n , j)  for(int i = n ; i >= j ; i--)
    #define ME(x , y) memset(x , y , sizeof(x))
    //ll lcm(ll a , ll b){return a*b/gcd(a,b);}
    //ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
    //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
    //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
    #define INF  0x3f3f3f3f
    #define PI acos(-1)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    #define pb push_back
    #define mp make_pair
    #define cin(x) scanf("%lld" , &x);
    using namespace std;
    const int N = 1e7+9;
    const int maxn = 2e3+9;
    const double esp = 1e-6;
    int dp[maxn][maxn];
    int a[maxn];
    
    void solve(){
        int n , h , l , r ;
        cin >> n >> h >> l >> r ;
        rep(i , 1 , n){
            cin >> a[i];
        }
        rep(i , 0 , n){
            rep(j , 0 , h-1){
                dp[i][j] = -1 ;
            }
        }
        dp[0][0] = 0;
        for(int i = 1 ; i <= n ; i++){
            for(int j = 0 ; j < h ; j++){
                if(dp[i-1][j] != -1){
                    int k = (j + a[i])%h;
                    if(k >= l && k <= r){
                        dp[i][k] = max(dp[i][k] , dp[i-1][j] + 1) ;
                    }else{
                        dp[i][k] = max(dp[i][k] , dp[i-1][j]);
                    }
                    k = (j + a[i] - 1) % h ;
                    if(k >= l && k <= r){
                        dp[i][k] = max(dp[i][k] , dp[i-1][j] + 1);
                    }else{
                        dp[i][k] = max(dp[i][k] , dp[i-1][j]);
                    }
                }
            }
        }
        int ans = -1 ;
        for(int i = 0 ; i < h ; i++){
            ans = max(ans , dp[n][i]);
        }
        cout << ans << endl;
    
    }
    
    signed main()
    {
        //ios::sync_with_stdio(false);
        //cin.tie(0); cout.tie(0);
        //int t ;
        //cin >> t ;
        //while(t--){
            solve();
        //}
    }
    

     

    记忆化递归:

    #include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <stdio.h>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <string.h>
    #include <vector>
    typedef long long ll ;
    #define int ll
    #define mod 1000000007
    #define gcd __gcd
    #define rep(i , j , n) for(int i = j ; i <= n ; i++)
    #define red(i , n , j)  for(int i = n ; i >= j ; i--)
    #define ME(x , y) memset(x , y , sizeof(x))
    //ll lcm(ll a , ll b){return a*b/gcd(a,b);}
    ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
    //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
    //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
    #define SC scanf
    #define INF  0x3f3f3f3f
    #define PI acos(-1)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    using namespace std;
    const int N = 1e6+100;
    const int maxn = 2e3+9;
    int dp[maxn][maxn];
    int a[maxn];
    int n , h , l , r;
    int dfs(int pos , int sum = 0){
        if(pos > n) return 0;
        int& ret = dp[pos][sum];
        if(ret != -1) return dp[pos][sum];
        int c = (sum + a[pos] - 1) % h ;
        int d = (sum + a[pos]) % h ;
        int a = dfs(pos + 1 , c) + (c >= l && c <= r) ;
        int b = dfs(pos + 1 , d) + (d >= l && d <= r) ;
        return dp[pos][sum] = max(a , b) ;
    }
    
    void solve(){
    
        cin >> n >> h >> l >> r ;
        rep(i , 1 ,n){
            cin >> a[i];
        }
        rep(i , 0 , n){
            rep(j , 0 , h-1){
                dp[i][j] = -1 ;
            }
        }
        cout << dfs(1) << endl;
    }
    
    signed main()
    {
        //ios::sync_with_stdio(false);
        //cin.tie(0); cout.tie(0);
        //int t ;
        //cin >> t ;
        //while(t--){
            solve();
        //}
    }
    

      

  • 相关阅读:
    Go语言的运算符
    Nginx基本安全优化
    在LNMP环境中部署一个blog服务程序
    PHP缓存加速器
    Go语言基础语法
    Go语言数据类型
    Go语言变量
    Go语言常量
    Go语言结构
    LNMP之PHP安装
  • 原文地址:https://www.cnblogs.com/nonames/p/12484091.html
Copyright © 2020-2023  润新知