UVa 133 - The Dole Queue
2016-08-02 09:46
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The Dole Queue
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~~ represents a space.
一个小破题目,烦死我了,一开始自己写没有考虑重复出现的字母
写的自己都恶心的代码,顺着思路来的,随便看看就好
书上用函数,看不懂,懒得看
遍历循环找到输出结果,然后用0取代
AC
#include <cstdio> #define maxn 25 int a[maxn]; int main() { int n,k,m; while(scanf("%d%d%d",&n,&k,&m)) { if(n==0&&k==0&&m==0) break; for(int i=1; i<=n; i++) a[i]=i; int left=n; int wow=0,xox=n+1; while(left) { int cnt=1;//计数,K次输出一个数 int flag=1;//flag输出一个数 int x,y; int i=0,j=0; for(i=wow+1; cnt<=k; i++)//每次都是输出的下一个开始 { if(i==n+1) i=1; if(cnt==k) { x=a[i]; wow=i;//将输出的数位置进行标记 } if(a[i]!=0) cnt++; } for(j=xox-1; flag<=m; j--) { if(j==-1) j=n; if(flag==m) { y=a[j]; xox=j; } if(a[j]!=0) flag++; } a[wow]=0; a[xox]=0; if(x==y) { printf("%3d",x); left-=1; } else { printf("%3d%3d",x,y); left-=2; } if(left) printf(","); else printf("\n"); } } return 0;
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