zoukankan      html  css  js  c++  java
  • Codeforces 792B. Counting-out Rhyme

    B. Counting-out Rhyme
    time limit per test:
    1 second
    memory limit per test:
    256 megabytes
    input:
    standard input
    output:
    standard output

    n children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to n. In the beginning, the first child is considered the leader. The game is played in k steps. In the i-th step the leader counts out ai people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.

    For example, if there are children with numbers [8, 10, 13, 14, 16] currently in the circle, the leader is child 13 and ai = 12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.

    You have to write a program which prints the number of the child to be eliminated on every step.

    Input

    The first line contains two integer numbers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ n - 1).

    The next line contains k integer numbers a1, a2, ..., ak (1 ≤ ai ≤ 109).

    Output

    Print k numbers, the i-th one corresponds to the number of child to be eliminated at the i-th step.

    Examples
    input
    7 5
    10 4 11 4 1
    output
    4 2 5 6 1 
    input
    3 2
    2 5
    output
    3 2 
    Note

    Let's consider first example:

    • In the first step child 4 is eliminated, child 5 becomes the leader.
    • In the second step child 2 is eliminated, child 3 becomes the leader.
    • In the third step child 5 is eliminated, child 6 becomes the leader.
    • In the fourth step child 6 is eliminated, child 7 becomes the leader.
    • In the final step child 1 is eliminated, child 3 becomes the leader.

    题目链接:http://codeforces.com/problemset/problem/792/B

    提意:编号1~n的n个人人围成一个圈。k此操作,每次操作为将执行者后面的第a个人剔除出去,下一次的执行者为剔除出去的后一个人,第一次操作的执行者编号为1。

    思路:模拟。将a对cou取模,cou为每次操作后的剩余人数。

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int sign[200];
    int main()
    {
        int n,k;
        int a;
        scanf("%d%d",&n,&k);
        int cou=n,start=0;
        memset(sign,0,sizeof(sign));
        for(int i=1; i<=k; i++)
        {
            scanf("%d",&a);
            a=a%cou;
            int t=0;
            while(t<=a)
            {
                start++;
                if(start>n) start-=n;
                if(sign[start]) continue;
                if(t==a) {sign[start]=1;cout<<start<<" ";}
                t++;
            }
            cou--;
        }
        return 0;
    }
    View Code
    I am a slow walker,but I never walk backwards.
  • 相关阅读:
    程序员修神之路--容器技术为什么会这么流行
    程序员修神之路--kubernetes是微服务发展的必然产物
    程序员修神之路--有状态的服务其实可以做更多的事情
    程序员修神之路--要想做好微服务架构,并非易事!
    程序员修神之路--为什么有了SOA,我们还用微服务?
    程序员过关斩将--数据库的乐观锁和悲观锁并非真实的锁
    程序员修神之路--设计一套RPC框架并非易事
    计算机的诞生和简史
    记一次Linux修改MySQL配置不生效的问题
    为什么大多数公司都不重视技术?
  • 原文地址:https://www.cnblogs.com/GeekZRF/p/6690746.html
Copyright © 2011-2022 走看看