1.完美数

暴力

#include <iostream>
#include <vector>

using namespace std;

bool perfect(long long res) {
    if (res <= 10) {
        return true;
    }
    else if (res > 10 && res <= 100 && res % 10 == 0) {
        return true;
    }
    else if (res > 100 && res <= 1000 && res % 100 == 0) {
        return true;
    }
    else if (res > 1000 && res <= 10000 && res % 1000 == 0) {
        return true;
    }
    else if (res > 10000 && res <= 100000 && res % 10000 == 0) {
        return true;
    }
    else if (res > 100000 && res <= 1000000 && res % 100000 == 0) {
        return true;
    }
    else if (res > 1000000 && res <= 10000000 && res % 1000000 == 0) {
        return true;
    }
    else if (res > 10000000 && res <= 100000000 && res % 10000000 == 0) {
        return true;
    }
    else if (res > 100000000 && res <= 1000000000 && res % 100000000 == 0) {
        return true;
    }
    return false;
}

int main() {
    int n;
    cin >> n;
    vector<int> nums(n);
    for (int i = 0; i < n; i++) {
        cin >> nums[i];
    }
    int count = 0;
    
    for (int left = 0; left < n - 1; left++) {
        for (int right = left + 1; right < n; right++) {
            long long product = (long long)nums[left] * (long long)nums[right];
            if (perfect(product)) {
                count++;
            }
        }
    }

    cout << count << endl;
    return 0;
}

补充gpt

#include <iostream>
#include <vector>

using namespace std;

bool perfect(long long res) {
    bool has_non_zero_digit = false;
    while (res > 0) {
        int digit = res % 10;
        if (digit != 0) {
            if (has_non_zero_digit) {
                return false;  // 如果已经有一个非零数字，且当前又有一个非零数字，则不是完美数
            }
            has_non_zero_digit = true;
        }
        res /= 10;
    }
    return has_non_zero_digit;
}

int main() {
    int n;
    cin >> n;
    vector<int> nums(n);
    for (int i = 0; i < n; i++) {
        cin >> nums[i];
    }
    int count = 0;
    
    for (int left = 0; left < n - 1; left++) {
        for (int right = left + 1; right < n; right++) {
            long long product = (long long)nums[left] * (long long)nums[right];
            if (perfect(product)) {
                count++;
            }
        }
    }

    cout << count << endl;
    return 0;
}

2.可爱串

不会

3.好二叉树

奇数符合，偶数排除，0的时候有1种

#include <iostream>
#include <vector>

using namespace std;

const int MOD = 1e9 + 7;

int main() {
    int n;
    cin >> n;

    vector<long long> dp(n + 1, 0);
    dp[0] = 1;  
    
    for (int i = 1; i <= n; i += 2) {
        for (int j = 0; j <= i - 1; j++) {
            dp[i] = (dp[i] + dp[j] * dp[i - 1 - j]) % MOD;
        }
    }

    cout << dp[n] << endl;
    return 0;
}