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  • POJ2657Comfort(扩展欧几里得基础)

    A game-board consists of N fields placed around a circle. Fields are successively numbered from1 to N clockwise. In some fields there may be obstacles. 

    Player starts on a field marked with number 1. His goal is to reach a given field marked with number Z. The only way of moving is a clockwise jump of length K. The only restriction is that the fields the player may jump to should not contain any obstacle. 

    For example, if N = 13, K = 3 and Z = 9, the player can jump across the fields 1, 4, 7, 10, 13, 3, 6 and 9, reaching his goal under condition that none of these fields is occupied by an obstacle. 

    Your task is to write a program that finds the smallest possible number K. 

    Input

    First line of the input consists of integers N, Z and M, 2 <= N <= 1000, 2 <= Z <= N, 0 <= M <= N - 2. N represents number of fields on the game-board and Z is a given goal-field. 

    Next line consists of M different integers that represent marks of fields having an obstacle. It is confirmed that fields marked 1 and Z do not contain an obstacle.

    Output

    Output a line contains the requested number K described above.

    Sample Input

    9 7 2
    2 3
    

    Sample Output

    3

    题意:

    现在我们要找一个最小的a,使得男主从一号点出发,每次跳a格,中途不遇到障碍,最终走到Z点。

    思路:

    有方程 1+ax=ny+z。 最小的a需要保证x正整数解,且不存在s<x,使得1+as=ny+障碍的位置。

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    using namespace std;
    int r[1010],n,z,m;
    void Ex_gcd(int a,int b,int &d,int &x,int &y)
    {
        if(b==0){x=1;y=0;d=a;return ;}
        Ex_gcd(b,a%b,d,y,x); y-=a/b*x;
    }
    bool check(int a)
    {
        int d,x,y,X;
        Ex_gcd(a,n,d,X,y);
        if((z-1)%d!=0) return false;
        X=(((z-1)/d*X)%(n/d)+n/d)%(n/d);
        for(int i=1;i<=m;i++){
            Ex_gcd(a,n,d,x,y);
            if((r[i]-1)%d!=0) continue;
            x=(((r[i]-1)/d*x)%(n/d)+n/d)%(n/d);
            if(x<=X) return false;
        } return true;
    }
    int main()
    {
        while(~scanf("%d%d%d",&n,&z,&m)){
            for(int i=1;i<=m;i++) scanf("%d",&r[i]);
            for(int i=1;i<n;i++) 
              if(check(i)){
                 printf("%d
    ",i); break;
              }
        } return 0;
    }
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  • 原文地址:https://www.cnblogs.com/hua-dong/p/8081366.html
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