Volume 1. Maths - Misc

113 - Power of Cryptography

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        BigInteger num1, num2;
        int n;
        int kc = 1000000000;
        int l, r, mid, t;
        Scanner in = new Scanner(System.in);
        while (in.hasNext())
        {
            n = in.nextInt();
            num1 = in.nextBigInteger();
            l = 0;
            r = kc;
            while (true)//二分
            {
                mid = (l + r) >> 1;
                num2 = BigInteger.valueOf(mid);
                num2 = num2.pow(n);
                t = num2.compareTo(num1);
                if (t == 0)
                {
                    System.out.println(mid);
                    break;
                }
                else
                    if (t > 0)
                        r = mid - 1;
                    else
                        l = mid + 1;
            }
        }
    }

}

10161 Ant on a Chessboard

#include <iostream>
#include <fstream>
#include <cstring>
#include <string>
#include <algorithm>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>

using namespace std;

int main()
{
    long long int n, nt, mid, x, y;
    while (cin>>n, n)
    {
        x = y = nt = ceil(sqrt(n));
        mid = (nt * nt + (nt - ) * (nt - ) + ) >> ;/**< long long 防止mid超了 */
        //cout<<mid<<endl;
        if (n != mid)
        {
            if (n > mid)
            {
                if (nt & )/**< n为奇数， 从左到右，再下 在左边*/
                {
                    x = x - n + mid;
                }
                else/**< 从下到上再到左 在下边*/
                {
                    y = y - n + mid;
                }
            }
            else
            {
                if (nt & )/**< n为奇数， 从左到右，再下 在下边*/
                {
                    y = y + n - mid;
                }
                else/**< 从下到上再到左 在左边*/
                {
                    x = x + n - mid;
                }
            }
        }
        cout<<x<<' '<<y<<endl;
    }
    return ;
}

253 - Cube painting

#include <iostream>
#include <fstream>
#include <cstring>
#include <string>
#include <algorithm>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>

using namespace std;
char s[], ts[];
int a[][] =/**< 枚举，总共24种情况，
在枚举中可以发现(1, 6)(2, 5)(3, 4)两两一组，
当其中一组换位置时，另外两组中必须只有一组需要换位置，
或者但两组互换时，其中一组也要换位置 */
{
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , , ,
    , , , , ,
};
bool fun()
{
    int i, j;
    for (i = ; i < ; i++)
    {
        for (j = ; j < ; j++)
            if (s[a[][j]] != ts[a[i][j]])
                break;
        if (j == )
            return true;
    }
    return false;
}
int main()
{
    int i = ;
    while (cin>>s[i++])
    {
        for (; i < ; i++)
            cin>>s[i];
        s[i] = '\0';
        for (i = ; i < ; i++)
            cin>>ts[i];
        ts[i] = '\0';
       cout<<(fun() ? "TRUE" : "FALSE")<<endl;
       if (cin.get() == EOF)
            break;
       i = ;
    }
    return ;
}

10025 - The ? 1 ? 2 ? ... ? n = k problem

 /**< 思路：
 先记f(n) = (n*(n+1))/2;
 现将k转换为正数，求k对应于f(n)的最小上界n，
 从n开始，
 第一步:如果f(n) - k这个数是偶数,则输出n，否则
 n++，继续第一步。

 i = f(n) - k>=0肯定成立，
 如果i为要使得f(n) = ?1?2?...?n = k;
 就要在f(n)中将i的一半为正，一半为负即可。所以i必定为偶数。
  */
 #include <iostream>
 #include <cmath>
 using namespace std;

 int main()
 {
     int n, k, i;
     cin>>n;
     while (n--)
     {
         cin>>k;
         if (k == )//为0输出3
         {
             i = ;
         }
         else
         {
             if (k < )
             {
                 k = -k;
             }
             i = sqrt(k * );
             if (((i * (i + )) /  - k) < )
                 i++;
             while (((i * (i + )) /  - k) & )
             {
                 i++;
             }
         }

         cout<<i<<endl;
         if (n)
         {
             cout<<endl;
         }
     }
     return ;
 }

591 - Box of Bricks

#include <iostream>
#include <cmath>
using namespace std;
int arr[];
int main()
{
    int n, avg, s, t = ;
    while (cin>>n, n)
    {
        s = avg = ;
        for (int i = ; i < n; i++)
        {
            cin>>arr[i];
            avg += arr[i];
        }
        avg /= n;
        for (int i = ; i < n; i++)
        {
            s += abs(arr[i] - avg);
        }
        cout<<"Set #"<<t++<<endl;
        cout<<"The minimum number of moves is "<<s / <<"."<<endl<<endl;
    }
    return ;
}

621 - Secret Research

//这样就过了
#include <iostream>
#include <string>
using namespace std;
string s;
int main()
{
    int n, len;
    cin>>n;
    char r;
    while(n--)
    {
        cin>>s;
        len = s.size() - ;
        if (len <= )
        {
            r = '+';
        }
        else
            if (s[len] == '' && s[len - ] == '')
            {
                r = '-';
            }
            else
                if (s[len] == '' && s[] == '')
                {
                    r = '*';
                }
                else
                {
                    r = '?';
                }
        cout<<r<<endl;
    }
    return ;
}

//这样始终过不了，实在是想不明白
#include <iostream>
#include <string>
using namespace std;
string s;
int main()
{
    int n, len;
    cin>>n;
    char r;
    while(n--)
    {
        cin>>s;
        len = s.size() - ;
        if (len <= )
        {
            r = '+';
        }
        else
            if (s[len] == '')
            {
                r = '-';
            }
            else
                if (s[len] == '')
                {
                    r = '*';
                }
                else
                {
                    r = '?';
                }
        cout<<r<<endl;
    }
    return ;
}

846 - Steps

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int t, x, y, ans;
    cin>>t;
    while(t--)
    {
        cin>>x>>y;
        ans = ;
        x = abs(x - y);
        if (x <= )
        {
            ans = x;
        }
        else
        {
            y = sqrt(x);
            while (x - ( + y) * y /  < (y - ) * y / )
            {
                y--;
            }
            while (x - ( + y) * y /  - (y - ) * y /  > )
            {
                ans++;
                x -= y;
            }
            if (x - ( + y) * y /  - (y - ) * y /  == )
            {
                ans++;
            }
            ans +=  * y - ;
        }
        cout<<ans<<endl;
    }
    return ;
}

573 - The Snail

#include <iostream>

using namespace std;

int main()
{
    double h, u, d, f;
    int days;
    double dis, dre;
    while (cin>>h>>u>>d>>f, h)
    {
        days = ;
        dis = ;
        dre = u * f / ;
        while (true)
        {
            dis += u;
            if (dis > h)
            {
                cout<<"success on day "<<days<<endl;
                break;
            }
            dis -= d;
            if (dis < )
            {
                cout<<"failure on day "<<days<<endl;
                break;
            }
            u -= dre;
            if (u < )
            {
                u = ;
            }
            days++;
        }
    }
    return ;
}

10499 - The Land of Justice

#include <iostream>

using namespace std;

int main()
{
    long long int n;
    while (cin>>n, n >= )
    {
        if (n <= )
            cout<<"0%"<<endl;
        else
            cout<< * n<<"%"<<endl;
    }
    return ;
}

10790 - How Many Points of Intersection?

/*
f(a, b)=f(a-1,b)+¡¾(a-1)*(b-1) + 0¡¿* b/2
f(a-1, b)=f(a-2,b)+¡¾(a-2)*(b-1) + 0¡¿* b/2
f(a-2, b)=f(a-3,b)+¡¾(a-3)*(b-1) + 0¡¿* b/2
f(2, b)=0+¡¾(b-1)¡¿* b/2
F(a,b) =(b-1)*b/2 *¡¾1+ ...+ (a-3) + (a-2)+(a-1)¡¿
=(b-1)*(b)/2 * a(a-1)/2
*/
#include <iostream>

using namespace std;

int main()
{
    long long a, b, t = , bb;
    while (cin>>a>>b, a | b)
    {
        if (b & )
        {
            bb = (b - ) / ;
        }
        else
        {
            bb = b / ;
            b = b - ;
        }
        cout<<"Case "<<t++<<": "<<a * (a - ) /  * b * bb<<endl;
    }
    return ;
}