演算法競賽入門經典(第二版)-劉汝佳-第四章 函式與遞迴 發放救濟金Uva133
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).
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).
10 4 3 0 0 0
4 8, 9 5, 3 1, 2 6, 10, 7
where represents a space.
#include<stdio.h>
#include<string.h>
int main()
{
int *p,*q,i,j,n,k,m,a[25]={},tot=0,t1,t2,k1,k2,case1,case2,c1=0,c2=0;
while(scanf("%d%d%d",&n,&k,&m)!=EOF&&n&&k&&m)
{
tot=0;
memset(a,0,sizeof(a));
for(i=n,j=1;i>=1;i--,j++)
{
a[i]=j;
}
q=a+1;p=a+n;
t1=*p;t2=*q;
while(tot!=n)
{ k1=k;
k2=m;
c1=c2=0;
if(*p==0)
c1--;
while(c1!=k-1)
{
p--;
if(p<=a)
{
p=a+n;
}
if(*p!=0)
{
c1++;
}
t1=*p;
}
if(*q==0)
c2--;
while(c2!=m-1)
{
q++;
if(q>=a+n+1)
{
q=a+1;
}
if(*q!=0)
{
c2++;
}
t2=*q;
}
if(*p==*q)
{
tot++;
if(tot==n-1&&tot==n)
printf("%3d",*p);
else
printf("%3d",*p);
*p=*q=0;
p--;
if(p<=a)
p=a+n;
q++;
if(q>=a+n)
q=a+1;
}
else
{
tot+=2;
if(tot==n-1||tot==n)
printf("%3d%3d",*p,*q);
else
printf("%3d%3d",*p,*q);
*p=*q=0;
p--;
if(p<=a)
p=a+n;
q++;
if(q>=a+n)
q=a+1;
}
if(tot!=n)
printf(",");
}
if(tot!=n)
printf(",");
printf("\n");
}
}