PTA2021 跨年挑战赛部分题解

7-1 压岁钱

不用说

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    int a, b, c, d;

    cin >> a >> b >> c >> d;

    int sum = a + b + c + d;

    cout << sum;

    return 0;

}

7-2 射击成绩

微米转毫米按环判断。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    double n;

    cin >> n;

    if(n <= 11500 / 2)

        cout<<10;

    else if(n <= 27500 / 2)

        cout<<9;

    else if(n <= 43500 / 2)

        cout<<8;

    else if(n <= 59500 / 2)

        cout<<7;

    else if(n <= 75500 / 2)

        cout<<6;

    else if(n <= 91500 / 2)

        cout<<5;

    else if(n <= 107500 / 2)

        cout<<4;

    else if(n <= 123500 / 2)

        cout<<3;

    else if(n <= 139500 / 2)

        cout<<2;

    else if(n <= 155500 / 2)

        cout<<1;

    else cout<<0;

    return 0;

}

7-3 Cassels方程

不用说

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    int n;

    cin >> n;

    while (n--) {

        int x, y, z;

        cin >> x >> y >> z;

        if (x * x + y * y + z * z != 3 * x * y * z)

            cout << "No" << endl;

        else cout << "Yes" << endl;

    }

    return 0;

}

7-4 相生相克

根据题意相生相克的数字和判断，也可以直接从金到土的数字看另一个数字是啥判断相生还是相克，因为相生相克都是一对一的。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

void judeg(int a, int b) {

    int sum = a + b;

    if (sum == 3)

        cout << "1 ke 2\n";

    else if (sum == 7) {

        if (a == 2 || a == 5)

            cout << "2 ke 5\n";

        else cout << "3 ke 4\n";

    } else if (sum == 8)

        cout << "5 ke 3\n";

    else if (sum == 5) {

        if (a == 4 || a == 1)

            cout << "4 ke 1\n";

        else cout << "3 sheng 2\n";

    } else if (sum == 6) {

        if (a == 2 || a == 4) {

            cout << "2 sheng 4\n";

        } else cout << "5 sheng 1\n";

    } else if (sum == 9)

        cout << "4 sheng 5\n";

    else if (a == 1 || a == 3)

        cout << "1 sheng 3\n";

}

int main() {

    int n;

    cin >> n;

    while (n--) {

        int x, y;

        cin >> x >> y;

        judeg(x, y);

    }

    return 0;

}

7-5 7-6太菜了没过

7-5 整除阶乘

对于每个数，直接把n * n + 1对n!的各乘因子求余，最后判断n * n + 1是否变成1来输出结果，用f做是否有结果标记，如果没有就输出None。来自：csdn

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    int n, m, f = 0;

    cin >> n >> m;

    for (int i = n; i <= m; i++) {

        int sum = i * i + 1;

        for (int j = 2; j <= i; j++) {

            if (sum >= j && sum % j == 0)

                sum /= j;

            else if (sum < j && j % sum == 0)

                sum = 1;

        }

        if (sum == 1) {

            cout << i << endl;

            f = 1;

        }

    }

    if (!f)cout << "None";

    return 0;

}

7-7 打PTA

先判断最后一个字符是否是？，不是直接输出enen，是就从下标2开始判断此下标是否是字符A和前面两个字符是否是T和P，是的话就把flag f设为真，f默认为假，再根据f的真假输出结果。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    int n;

    cin >> n;

    for (int j = 0; j < n; j++) {

        string s;

        if (j == 0)

            getchar();

        getline(cin, s);

        int t = s.length();

        int f = 0;

        if (s[t - 1] != '?') {

            cout << "enen\n";

        } else {

            for (int i = 2; i < t; i++) {

                if (s[i] == 'A' && s[i - 1] == 'T' && s[i - 2] == 'P') {

                    f = 1;

                    break;

                }

            }

            if (f)

                cout << "Yes!\n";

            else cout << "No.\n";

        }

    }

    return 0;

}

7-8 完美对称

从头到尾开始判断区间是否对称，不对称头就顺移到下一位直到找到对称区间，当头等于第一位时就直接完整输出，不是时倒序输出头前面区间。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e9;

const int maxm = 1e5 + 5;

const ll inf = 2147483647;

using namespace std;

int main() {

    int n;

    cin >> n;

    vector<int> v(n + 1);

    for (int i = 1; i <= n; i++) {

        cin >> v[i];

    }

    for (int i = 1; i <= n; i++) {

        int k = n, j = i, f = 1;

        while (k >= j) {//这里判断是否对称

            if (v[k] != v[j]) {

                f = 0;

                break;

            }

            k--;

            j++;

        }

        if (f && i != 1) {

            cout << v[i - 1];

            for (int p = i - 2; p >= 1; p--) {

                cout << ' ' << v[p];

            }

            break;

        } else if (f) {

            cout << v[1];

            for (int p = 2; p <= n; p++) {

                cout << ' ' << v[p];

            }

            break;

        }

    }

    return 0;

}

PTA2021 跨年挑战赛部分题解的相关教程结束。