UV 133-The Dole Queue // simulation, linked list

In fact, this question is not difficult, that is, the output requirements are a little higher. As a result, I have learned a lot, resulting in a long and messy code, resulting in wa and pe.

Question Analysis: If you want to open an array and simulate a ring, you can directly use an array to simulate a circular linked list + Delete. OK. An error occurred while understanding the output. As a result, WAF performed 3 times and peat twice. Here is a small trap:

If the point to be deleted is exactly the point where another person starts counting again, move one digit forward or backward.

The following code is used:

#include<stdio.h>#define MAXN 22int queue[MAXN];int prio[MAXN],next[MAXN];int n,k,m,N;void initi(){    for(int i=1;i<=n;i++)    {        queue[n-i+1]=i;        if(i==1) prio[n-i+1]=1;        else prio[n-i+1]=n-i+2;        if(i==n) next[n-i+1]=n;        else next[n-i+1]=n-i;    }}int couter_clock(int begin){    for(int i=1;i<k;i++) begin=prio[begin];    return begin;}int clock(int begin){    for(int i=1;i<m;i++) begin=next[begin];    return begin;}void del(int begin){    prio[next[begin]]=prio[begin];    next[prio[begin]]=next[begin];    N--;}int main(){    while(scanf("%d%d%d",&n,&k,&m),n,k,m){    initi();    int begin1=1;    int begin2=n;    N=n;    while(true)    {        int ans1=couter_clock(begin1);        int ans2=clock(begin2);        begin1=prio[ans1];        begin2=next[ans2];        del(ans1);        if(ans1!=ans2) del(ans2);        if(begin1==ans2) begin1=prio[begin1];        if(begin2==ans1) begin2=next[begin2];        printf("%3d",ans1);        if(ans1!=ans2) printf("%3d",ans2);        if(N>0) printf(",");        else {printf("\n");break;}    }    }    return 0;}

Hey, although the question is simple, the coding capability is improved! Keep moving!

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