小数化分数2

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>

using namespace std;

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

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

    while (n -- ) {
        cin >> a;
        int pos = 0; // 不循环的位数
        int POS = 0; // 循环的位数
        bool flag = false; // 是否有循环

        for (int i = 2; i < a.size(); i ++ ) {

            if (!flag && a[i] >= '0' && a[i] <= '9')
                pos ++ ;

            if (a[i] == '(')
                flag = true;

            if (flag && a[i] >= '0' && a[i] <= '9')
                POS ++ ;
        }

        int T = 0;

        for (int i = 2; i <= pos + 1; i ++ )

            T = T * 10 + (a[i] - '0');

        if (!flag) {
            int x = (int)pow(10, pos);
            int y = gcd(T, x);
            cout << T / y << '/' << x / y << endl;
        } else {
            int ans = 0;

            for (int i = 2; i < a.size(); i ++ )

                if (a[i] >= '0' && a[i] <= '9')
                    ans = ans * 10 + (a[i] - '0');
            ans -= T;
            int x = (int)pow(10, pos);
            int X = (int)pow(10, a.size() - 4);
            X -= x;
            int y = gcd(ans, X);
            cout << ans / y << '/' << X / y << endl;
        }
    }

    return 0;
}

N!

#include <iostream>
#include <vector>

using namespace std;

vector<int> C;

void factor(int n)
{
    C.push_back(1); // 如果n == 1 ，答案数组为1 

    for (int i = 2; i <= n; i ++)  // n ！= 1 ， 执行递归 乘法代码 
    {
        int t = 0;  // t 是进位 
        for (int j = 0; j < C.size(); j ++)  // 遍历当前答案数组中的每一位 
        {
            C[j] = C[j] * i + t;
            t = C[j] / 10;
            C[j] %= 10;
        } 
        while (t)
        {
            C.push_back(t % 10);
            t /= 10;
        }
    }
}

int main()
{
    int n;
    while (cin >> n)
    {
        C.clear();    // 每次算阶乘之前要清空答案数组 
        factor(n);
        for (int i = C.size() - 1; i >= 0; i --) cout << C[i];
        cout << endl;
    }
    return 0;
}

Number Sequence

#include <iostream>
#include <cmath>

using namespace std;

typedef long long LL;

const int N = 35000;

LL a[N];
LL s[N];

void init() {
    a[1] = s[1] = 1;

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

        a[i] = a[i - 1] + (int)log10((double)i) + 1;
        s[i] = s[i - 1] + a[i];
    }
}

int solve(int n) {
    int i = 1;

    while (s[i] < n) {
        i ++ ;
    }

    int pos = n - s[i - 1];
    int len = 0;

    for (i = 1; len < pos; i ++ )

        len += (int)log10((double)i) + 1;
    return (i - 1) / (int)pow((double)10, len - pos) % 10;
}

int main() {
    init();
    int t;
    cin >> t;

    while (t -- ) {
        int n;
        cin >> n;
        cout << solve(n) << endl;
    }

    return 0;
}

Fibonacci

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int f[21] = {0, 1, 1};

void init() {
    for (int i = 3; i < 21; i ++ )

        f[i] = f[i - 1] + f[i - 2];
}

int main() {
    init();
    int n;

    while (cin >> n) {
        if (n <= 20) {
            cout << f[n] << endl;
            continue;
        } else {
            double temp = -0.5 * log(5.0) / log(10.0) + ((double)n) * log((sqrt(5.0) + 1.0) / 2.0) / log(10.0);
            temp -= floor(temp);
            temp = pow(10.0, temp);

            while (temp < 1000)
                temp *= 10;
            cout << (int)temp << endl;
        }
    }

    return 0;
}

Power of Cryptography

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    double n, p;

    while (cin >> n >> p) {
        cout << pow(p, 1.0 / n) << endl;
    }

    return 0;
}