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  • cf467B Fedor and New Game

    B. Fedor and New Game
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3».

    The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered form 1 to (m + 1). Types of soldiers are numbered from 0 to n - 1. Each player has an army. Army of the i-th player can be described by non-negative integer xi. Consider binary representation of xi: if the j-th bit of number xi equal to one, then the army of the i-th player has soldiers of the j-th type.

    Fedor is the (m + 1)-th player of the game. He assume that two players can become friends if their armies differ in at most k types of soldiers (in other words, binary representations of the corresponding numbers differ in at most k bits). Help Fedor and count how many players can become his friends.

    Input

    The first line contains three integers nmk (1 ≤ k ≤ n ≤ 20; 1 ≤ m ≤ 1000).

    The i-th of the next (m + 1) lines contains a single integer xi (1 ≤ xi ≤ 2n - 1), that describes the i-th player's army. We remind you that Fedor is the (m + 1)-th player.

    Output

    Print a single integer — the number of Fedor's potential friends.

    Sample test(s)
    input
    7 3 1
    8
    5
    111
    17
    
    output
    0
    
    input
    3 3 3
    1
    2
    3
    4
    
    output
    3

    sb题*2

    给m个数,转成二进制,问和给定的x的每一位异或和小等于k的有多少个

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    int n,m,k,ans,a[100100];
    int work(int x)
    {
        int res=0;
        for(int i=0;i<=21;i++)
         if((x&(1<<i))^(m&(1<<i)))res++;
        return res;
    }
    int main()
    {
        scanf("%d%d%d",&m,&n,&k);
        int x,y;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=1;i<=n;i++)
            if(work(a[i])<=k)ans++;
        printf("%d",ans);
        return 0;
    }
    

      

    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4035989.html
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