PAT甲级官网 刷题（4）

PAT1114 Family Property

  并查集，麻烦的是需要把人的编号映射到0-n，同时还需要把0-n逆映射到原来的编号。

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>

using namespace std;
const int MAXN = 10010;
const double eps = 1e-6;
//1.2.3映射到原来多的数字
int map1[MAXN];
//数字映射到1.2.3.。。。
int map2[MAXN];
int cnt[MAXN], pre[MAXN], sets[MAXN], sum[MAXN];
int n, num = 1;

struct node {
    
    
    int num, cnt;
    double sets, sum;

    node(int _num, int _cnt, double _sets, double _sum) : num(_num), cnt(_cnt), sets(_sets), sum(_sum) {
    
    };

    bool operator<(node &a) const {
    
    
        if (abs(sum - a.sum) > eps)
            return sum > a.sum;
        else
            return num < a.num;
    }
};

vector<node> ans;

int find(int a) {
    
    
    return a == pre[a] ? a : pre[a] = find(pre[a]);
}

int main() {
    
    
    int a, b, c, len, t, estate, sts;

    cin >> n;
    while (n--) {
    
    
        vector<int> arr;
        cin >> a >> b >> c >> len;
        arr.push_back(a);
        if (b != -1)
            arr.push_back(b);
        if (c != -1)
            arr.push_back(c);
        for (int i = 0; i < len; i++) {
    
    
            cin >> a;
            arr.push_back(a);
        }

        for (int i = 0; i < arr.size(); i++) {
    
    
        	//当前人之前没有遇到过，分配一个编号给他
            if (!map2[arr[i]]) {
    
    
                map2[arr[i]] = num;
                map1[num] = arr[i];
                cnt[num] = 1;
                pre[num] = num;
                ++num;
            }
            if (!i)
                continue;
            int aa = find(map2[arr[i - 1]]);
            int bb = find(map2[arr[i]]);
            if (aa != bb) {
    
    
            	//编号小的为根
                if (map1[aa] < map1[bb]) {
    
    
                    pre[bb] = aa;
                    cnt[aa] += cnt[bb];
                    sets[aa] += sets[bb];
                    sum[aa] += sum[bb];
                } else {
    
    
                    pre[aa] = bb;
                    cnt[bb] += cnt[aa];
                    sets[bb] += sets[aa];
                    sum[bb] += sum[aa];
                }
            }
        }
        a = find(map2[arr[0]]);
        cin >> sts >> estate;
        sets[a] += sts;
        sum[a] += estate;
    }

    for (int i = 1; i < num; i++)
        if (i == find(i))
            ans.push_back(node(map1[i], cnt[i], (double) sets[i] / cnt[i], (double) sum[i] / cnt[i]));

    cout << ans.size() << endl;

    sort(ans.begin(), ans.end());
    for (node &a:ans)
        printf("%04d %d %.3lf %.3lf\n", a.num, a.cnt, a.sets, a.sum);
    return 0;
}

---

PAT1145 Hashing - Average Search Time

  简单的哈希映射。

#include<iostream>

#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
const int MAXN = 100010;
int arr[MAXN];
int N, M, K;

bool check(int N) {
    
    
    for (int i = 2; i * i <= N; i++)
        if (N % i == 0)
            return false;
    return true;
}

void insert(int a) {
    
    
    int pos = a % N;
    for (int i = 0; i < N; i++) {
    
    
        if (arr[(pos + i * i) % N] == -1) {
    
    
            arr[(pos + i * i) % N] = a;
            return;
        }
    }
    cout << a << " cannot be inserted." << endl;
}

int main() {
    
    
    ac
    int a;

    for (int i = 0; i < MAXN; i++)
        arr[i] = -1;

    cin >> N >> M >> K;
    while (!check(N))
        ++N;

    while (M--) {
    
    
        cin >> a;
        insert(a);
    }

    int cnt = 0;
    for (int j = 0; j < K; j++) {
    
    
        cin >> a;
        int pos = a % N, i = 0;
        //该位置是空位置但是不是该元素，也需要退出循环，代表这个数字不在序列中
        while (i < N && arr[(pos + i * i) % N] != a && arr[(pos + i * i) % N] != -1)
            ++i;
        cnt += i + 1;
    }
    printf("%.1lf", (double) cnt / K);
    return 0;
}

---

PAT1136 A Delayed Palindrome

  简单的高精度运算。

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

bool check(const string &a) {
    
    
    for (int i = 0; i < a.length() / 2; i++)
        if (a[i] != a[a.length() - 1 - i])
            return false;
    return true;
}

void add(string &a, string &b) {
    
    
    int carry = 0;

    for (int i = a.length() - 1; i >= 0; i--) {
    
    
        a[i] += b[i] - '0' + carry;
        carry = (a[i] - '0') / 10;
        a[i] = (a[i] - '0') % 10 + '0';
    }
    if (carry)
        a = '1' + a;
}

int main() {
    
    
    string a, b;

    cin >> a;
    for (int i = 0; i < 10; i++) {
    
    
        if (check(a)) {
    
    
            cout << a << " is a palindromic number.";
            return 0;
        }
        b = a;
        reverse(b.begin(), b.end());
        cout << a << " + " << b + " = ";
        add(a, b);
        cout << a << endl;
    }
    if (check(a))
        cout << a << " is a palindromic number.";
    else
        cout << "Not found in 10 iterations.";
    return 0;
}

---