uvaoj-133:约瑟夫环的模拟
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In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counter-clockwise up to N (who will be standing on 1's left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.
Input
Write a program that will successively read in (in that order) the three numbers (N, k and m; k, m > 0, 0 < N < 20) and determine the order in which the applicants are sent off for retraining. Each set of three numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).
Output
For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counter-clockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).
Sample input
10 4 30 0 0
Sample output
4 8, 9 5, 3 1, 2 6, 10, 7
where represents a space.
题解:刘汝佳白书82页例题4-3,思路就是模拟运行,利用自定义函数可以是程序更加美观;
code:
#include <cstring>
#include <cmath>
#define maxn 25
int n,k,s[maxn],m;
int go(int start,int dir,int step);//声明函数;
int main()
{
while(~scanf("%d%d%d",&n,&k,&m)&&n)
{
int left=n;
for(int i=1; i<=20; i++)//将数据存入数组以便在循环的时候跳过;
s[i]=i;
int a=n;
int b=1;
while(left)
{
a=go(a,1,k);//三个变量分别代表:初始位置、方向、步幅;
b=go(b,-1,m);
printf("%3d",a);
left--;
if(b!=a)
{
printf("%3d",b);
left--;
}
s[a]=s[b]=0;//将已经数过的人变为零避免下次循环的时候再次数发生错误;
if(left)
printf(",");
}
printf("\n");
}
return 0;
}
int go(int start,int dir,int step)//A,B两个人的方向正好相反,可以用正负1分别代表方向;
{
while(step--)//每次只数一个,数step次;
{
do
{
start=(start+n-1+dir)%n+1;//这里加一减一来避免出现start为零的情况;
}
while(s[start]==0);//do-while语句是先运行再判断,利用do-while将圈内已经数过的人跳过;
};
return start;
}
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