133 - The Dole Queue
2016-05-05 14:14
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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 3
0 0 0
Sample output
![](https://uva.onlinejudge.org/external/1/133img1.gif)
4
![](https://uva.onlinejudge.org/external/1/133img1.gif)
8,
![](https://uva.onlinejudge.org/external/1/133img1.gif)
9
![](https://uva.onlinejudge.org/external/1/133img1.gif)
5,
![](https://uva.onlinejudge.org/external/1/133img1.gif)
3
![](https://uva.onlinejudge.org/external/1/133img1.gif)
1,
![](https://uva.onlinejudge.org/external/1/133img1.gif)
2
![](https://uva.onlinejudge.org/external/1/133img1.gif)
6,
![](https://uva.onlinejudge.org/external/1/133img1.gif)
10,
![](https://uva.onlinejudge.org/external/1/133img1.gif)
7
where
![](https://uva.onlinejudge.org/external/1/133img1.gif)
represents a space.
n(n<20)个人站成一圈,逆时针编号为1~n。有两个官员,A从1开始逆时针数,B从n开始顺时针数。在每一轮中,官员A数k个就停下来,官员B数m个就停下来(注意有可能两个官员停在同一个人上)。接下来被官员选中的人(1个或者2个)离开队伍。输入n,k,m输出每轮里被选中的人的编号(如果有两个人,先输出被A选中的)。例如,n=10,k=4,m=3,输出为4 8, 9 5, 3 1, 2 6, 10, 7。注意:输出的每个数应当恰好占3列。
#include <stdio.h> #include <string.h> #define maxNum 25 // N为人数,k为军官1数的间隔人数 // m为军官2数的间隔人数 int N,k,m; // 标记数组 int mark[maxNum]; // 剩余人数 int num; // 军官1和军官2的当前位置 int kpos; int mpos; void choice() { int kt = k - 1; int mt = m - 1; while(kt || mark[kpos]) { if(!mark[kpos]) { kt--; } kpos++; if(kpos >= N + 1) { kpos = 1; } } while(mt || mark[mpos]) { if(!mark[mpos]) { mt--; } mpos--; if(mpos <= 0) { mpos = N; } } if (kpos == mpos && !mark[kpos]) { printf(" %c%c", (kpos) / 10 ? (kpos) / 10 + '0' : ' ', (kpos) % 10 + '0'); mark[kpos] = 1; num--; } else if (kpos != mpos) { if(!mark[kpos]) { printf(" %c%c", (kpos) / 10 ? (kpos) / 10 + '0' : ' ', (kpos) % 10 + '0'); mark[kpos] = 1; num--; } if(!mark[mpos]) { printf(" %c%c", (mpos) / 10 ? (mpos) / 10 + '0' : ' ', (mpos) % 10 + '0'); mark[mpos] = 1; num--; } } if(num) { printf(","); } else { printf("\n"); } } int main() { while(scanf("%d%d%d", &N, &k, &m) == 3 && N != 0) { // 每回合都重置标记数组 memset(mark, 0, sizeof(mark)); kpos = 1; mpos = N; num = N; while(num) { choice(); } } return 0; }
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