[hihocoder][Offer收割]编程练习赛46

AEIOU

选出的子串中由AEI构成的子串和由OU构成的子串之间并没有什么关系，分别算出最长的加起来。

#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);
}

string s;
int dp[];
int p[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    cin >> s;
    int n = s.size();
    int max1 = , max2 = ;
    memset(dp, , sizeof(dp));
    for (int i = ; i < ; i++) p[i] = -;
    for (int i = ; i < n; i++) {
        if (s[i] == 'a') {
            dp[i] = ;
            if (p['a'] != -) dp[i] = max(dp[i], dp[p['a']] + );
            if (dp[i] > dp[p['a']]) p['a'] = i;
        }
        if (s[i] == 'e') {
            dp[i] = ;
            if (p['a'] != -) dp[i] = max(dp[i], dp[p['a']] + );
            if (p['e'] != -) dp[i] = max(dp[i], dp[p['e']] + );
            if (dp[i] > dp[p['e']]) p['e'] = i;
        }
        if (s[i] == 'i') {
            dp[i] = ;
            if (p['a'] != -) dp[i] = max(dp[i], dp[p['a']] + );
            if (p['e'] != -) dp[i] = max(dp[i], dp[p['e']] + );
            if (p['i'] != -) dp[i] = max(dp[i], dp[p['i']] + );
            if (dp[i] > dp[p['i']]) p['i'] = i;
        }
        max1 = max(max1, dp[i]);
    }
    memset(dp, , sizeof(dp));
    for (int i = ; i < ; i++) p[i] = -;
    for (int i = ; i < n; i++) {
        if (s[i] == 'o') {
            dp[i] = ;
            if (p['o'] != -) dp[i] = max(dp[i], dp[p['o']] + );
            if (dp[i] > dp[p['o']]) p['o'] = i;
        }
        if (s[i] == 'u') {
            dp[i] = ;
            if (p['o'] != -) dp[i] = max(dp[i], dp[p['o']] + );
            if (p['u'] != -) dp[i] = max(dp[i], dp[p['u']] + );
            if (dp[i] > dp[p['u']]) p['u'] = i;
        }
        max2 = max(max2, dp[i]);
    }
    cout << max1 + max2 << endl;
    return ;
}

数字游戏

这题看别人的代码貌似跟全排列有关，但是没想通关系在哪。

我的解法是这样的：首先加减和交换的顺序并不会影响操作的步数，所以可以看做是先进行若干步交换，然后依次把每一位加减为B的对应位数字。以A为初始状态进行bfs，然后卡时，就过了、

#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);
}

char A[], B[];
int cost[][];
lint p[];
int n;
struct state {
    int step;
    char num[];
    bool operator < (state s) const {
        if (step == s.step) return false;
        else return step > s.step;
    }
};
lint calc(state s) {
    lint rtn = ;
    for (int i = ; i < n; i++) rtn += s.num[i] * p[n - i - ];
    return rtn;
}
map<lint, int> mp;
priority_queue<state> q;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    scanf("%d%s%s", &n, A, B);
    for (int i = ; i < n; i++) {
        A[i] -= '';
        B[i] -= '';
    }
    p[] = ;
    for (int i = ; i < ; i++) p[i] = p[i - ] * ;
    for (int i = ; i < ; i++) {
        for (int j = ; j < ; j++) {
            if (i == j) cost[i][j] = ;
            if (i > j) cost[i][j] = cost[j][i];
            if (i < j) cost[i][j] = min(j - i, i +  - j);
        }
    }
    lint b = ;
    for (int i = ; i < n; i++) b += B[i] * p[n - i - ];
    state s, u;
    s.step = ;
    memcpy(s.num, A, sizeof(A));
    q.push(s);
    mp[calc(s)] = ;
    int ans = , iter = ;
    while (!q.empty()) {
        s = q.top();
        q.pop();
        int step = s.step;
        for (int i = ; i < n; i++) step += cost[s.num[i]][B[i]];
        ans = min(ans, step);
        iter++;
        if (iter > ) break;
        for (int i = ; i < n; i++) {
            for (int j = i + ; j < n; j++) {
                u = s;
                char t = u.num[i];
                u.num[i] = u.num[j];
                u.num[j] = t;
                u.step++;
                lint c = calc(u);
                if (mp.find(c) != mp.end()) continue;
                else {
                    mp[c] = ;
                    q.push(u);
                }
            }
        }
    }
    cout << ans << endl;
    return ;
}

第K小分数

前K个分数显然由每个质数构成的分数的前连续若干项构成，由于最多有1000个质数，所以最多有1000个最后一项，所以第K个分数一定是这最多1000个最后一项中的一个。可以用二分法来确定每种分母的最后一项并计算其排名，若其排名为K，则找到了题目要求的分数。

#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 p[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n;
    lint k;
    cin >> n >> k;
    for (int i = ; i < n; i++) cin >> p[i];
    for (int i = ; i < n; i++) {
        lint l = , r = p[i], mid;
        while (l +  < r) {
            mid = (l + r) >> ;
            lint tmp = ;
            for (int j = ; j < n; j++) tmp += mid * p[j] / p[i];
            if (tmp == k) cout << mid << '/' << p[i] << endl;
            if (tmp <= k) l = mid;
            else r = mid;
        }
    }
    return ;
}

逆序异或和

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = ;

int n, a[N];

struct BIT {
    int c[N];
    void init() {
        memset(c, , sizeof(c));
    }
    void insert(int x) {
        for (; x < N; x += x & -x) c[x] ++;
    }
    int count(int x, int res=) {
        for (; x; x ^= x & -x) res += c[x];
        return res;
    }
}bit[], num;

int main() {
    scanf("%d", &n);
    for (int i = ; i <= n; ++i) scanf("%d", a + i);
    num.init();
    for (int i = ; i < ; ++i) bit[i].init();
    ll ans = ;
    for (int i = n; i >= ; --i) {
        int tar = num.count(a[i] - );
        num.insert(a[i]);

        for (int j = ; j < ; ++j) {
            int _num =  << j;
            int _bit = a[i] & _num;
            int _bit_tar = bit[j].count(a[i] - );
            if (_bit) {
                ans += 1LL * _num * (tar - _bit_tar);
                bit[j].insert(a[i]);
            }
            else {
                ans += 1LL * _num * _bit_tar;
            }
        }
    }
    cout << ans << endl;
    return ;
}