[hihocoder][Offer收割]编程练习赛50

循环数组

计算a[i]的前缀和s[i]，计算l[i]为1~i-1中最小的s值，r[i]为i~n中最大的s值。

则a[i]~a[n]满足性质的条件为r[i]-s[i-1]>0，a[1]~a[i-1]满足性质的条件为l[i]+s[n]-s[i-1]>0

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

lint a[], s[], l[], r[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n;
    cin >> n;

    for(int i = ; i <= n; i++) cin >> a[i];

    s[] = a[];

    for(int i = ; i <= n; i++) s[i] = s[i - ] + a[i];

    l[] = , l[] = s[];

    for(int i = ; i <= n; i++) l[i] = min(l[i - ], s[i - ]);

    r[n] = s[n];

    for(int i = n - ; i >= ; i--) r[i] = min(s[i], r[i + ]);

    for(int i = ; i <= n; i++) {
        if((r[i] - s[i - ] > ) && (s[n] - s[i - ] + l[i] > )) {
            cout << i << endl;
            return ;
        }
    }

    cout << - << endl;
    return ;
}

座位问题

根据题目要求，对于一段区间为l~r的连续空座位，其中最优先的位置应为(l+r)/2。对于多段连续的空座位，选择其中长度最长的那一段，如果有多段都是最长的，选择开始位置最小的那一段。坐下后一段可能变为两段、一段或零段，用一个小根堆维护这些段，新来的直接插在堆顶的段里。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}
class segment {
public:
    int l, r;
    bool operator <(const segment & s) const {
        if(r - l < s.r - s.l) return true;

        if(r - l > s.r - s.l) return false;

        return l > s.l;
    }
};
priority_queue<segment> q;
int a[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n, m, k;
    cin >> n >> m >> k;

    for(int i = ; i <= m; i++) cin >> a[i];

    a[] = , a[m + ] = n + ;

    for(int i = ; i <= m + ; i++) {
        //a[i - 1] + 1, a[i] - 1
        if(a[i - ] +  <= a[i] - ) {
            segment s;
            s.l = a[i - ] + ;
            s.r = a[i] - ;
            q.push(s);
        }
    }

    for(int i = ; i < k; i++) {
        segment s = q.top(), tmp;
        q.pop();
        int x = (s.l + s.r) / ;
        cout << x << endl;

        if(s.l <= x - ) {
            tmp.l = s.l, tmp.r = x - ;
            q.push(tmp);
        }

        if(x +  <= s.r) {
            tmp.l = x + , tmp.r = s.r;
            q.push(tmp);
        }
    }

    return ;
}

末尾有最多0的乘积

显然0的个数只与每个数中2和5的因子个数有关。

动态规划：dp[i][j][k]表示从前i个中选出j个得到k个2时能得到的最多的5的个数，值为-1表示该状态不存在。

dp[i][j][k] = max(dp[i - 1][j][k], dp[i - 1][j - 1][k - b[i]] + c[i]);

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

int dp[][][], b[], c[];
lint a[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n, m;
    cin >> n >> m;

    for(int i = ; i <= n; i++) {
        cin >> a[i];
        b[i] = c[i] = ;

        while(a[i] %  == ) a[i] /= , b[i]++;

        while(a[i] %  == ) a[i] /= , c[i]++;
    }

    memset(dp, -, sizeof(dp));

    for(int i = ; i <= n; i++) dp[i][][] = ;

    for(int i = ; i <= n; i++) {
        for(int j = ; j <= i; j++) {
            for(int k = ; k >= ; k--) {
                dp[i][j][k] = dp[i - ][j][k];

                if(k >= b[i] && dp[i - ][j - ][k - b[i]] != -) dp[i][j][k] = max(dp[i][j][k], dp[i - ][j - ][k - b[i]] + c[i]);
            }
        }
    }

    int ans = ;

    for(int i = ; i < ; i++) if(dp[n][m][i] != -)ans = max(min(dp[n][m][i], i), ans);

    cout << ans << endl;
    return ;
}