pat乙级每日习题

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1：https://pintia.cn/problem-sets/994805260223102976/problems/994805325918486528

#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
    int n;cin >> n;
    int ans = 0;
    while(n != 1){
        if((n & 1) == 1){
            n = (n * 3 + 1) >> 1;
        }else{
            n >>= 1;
        }
        ++ans;
    }
    cout << ans << endl;
    return 0;
}

2：https://pintia.cn/problem-sets/994805260223102976/problems/994805320306507776

#include <bits/stdc++.h>
using namespace std;

const int N = 110;

int arr[N];
int n;

int main(int argc, char const *argv[])
{
    unordered_map<int, bool> hash;

    cin >> n;
    for(int i = 1;i <= n;++i){
        cin >> arr[i];//存进数组因为求答案要用
        int x = arr[i];
        //把x最终变成1的所有中间的数都置为true，代表出现过
        while(x != 1){
            if((x & 1) == 1) //&1 == 1 代表是奇数
                x = (3 * x + 1) >> 1;
            else 
                x >>= 1;
            hash[x] = true;
        }
    }

    vector<int> ans;
    //把数组中的数没有被覆盖过的也就是 false 的加进答案
    for(int i = 1;i <= n;++i)
        if(!hash[arr[i]])
            ans.push_back(arr[i]);    
    //以下代码完全是为了格式化输出
    sort(ans.begin(), ans.end(), greater<int>());
    for(int i = 0;i < ans.size()-1;++i)
        cout << ans[i] << " ";
    cout << ans[ans.size()-1];
    cout << endl;

    return 0;
}

3：https://pintia.cn/problem-sets/994805260223102976/problems/994805317546655744

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5;

int n;
bool state[N];

int main(int argc, char const *argv[])
{
    cin >> n;
    int ans = 0;
    int pre = -1;
    for(int i = 2;i <= n;++i){
        if(!state[i]){
            if(i - pre == 2) 
                ++ans;
            pre = i;
            for(int j = i + i; j <= n; j += i)
                state[j] = true;
        }
    }
    cout << ans << endl;
    return 0;
}

4:https://pintia.cn/problem-sets/994805260223102976/problems/994805296180871168

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+10;

struct Node{
    int val;
    int next;
}node[N];

int head;
int n, k;

int revers(int cur, int t){
    int pre = -1;
    while(cur != t){
        int tmp = node[cur].next;
        node[cur].next = pre;
        pre = cur;
        cur = tmp;
    }
    return pre;
}

int help(int h){
    if(h == -1) return h;
    int tail = h;
    for(int i = 0;i < k;++i){
        tail = node[tail].next;
        if(tail == -1) return h;
    }
    int newh = revers(h, tail);
    node[h].next = help(tail);
    return newh;
}

int main(){
    cin >> head >> n >> k;
    for(int i = 1;i <= n;++i){
        int x;cin >> x;
        cin >> node[x].val >> node[x].next;
    }
    int newhead = help(head);
    for(int i = newhead;i != -1;i = node[i].next)
        if(node[i].next == -1)
            printf("%05d %d %d
", i, node[i].val, node[i].next);
        else
            printf("%05d %d %05d
", i, node[i].val, node[i].next);
    return 0;
}

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