UVA 1394 POJ 3517 And Then There Was One (雙向循環鏈表和遞推)

這題用雙向循環鏈表的時間是O（NK），明顯是過不了的，但是爲了練練手，還是寫了一下

#include <cstdio>
#include <algorithm>
using namespace std;
struct List
{
	List* pre;
	int value;
	List* next;
};
int main(){
	int i,j;

	int n,k,m;
	while(scanf("%d %d %d",&n,&k,&m)){
		List *list=new List;
		if(k==0 && m==0 && n==0)
			break;
		list->value=1;
		list->next=list;
		list->pre=list;
		List *start=list;
		for(i=2;i<=n;i++){
			List *a=new List;
			a->value=i;
			a->pre=list->pre;
			a->next=list;
			list->pre->next= a;
			list->pre=a;
			if(i==m)
				start=a;
		}
		if(n>1){
			start->pre->next=start->next;
			start->next->pre=start->pre;
			List *t=start->next;
			printf("%d\n",start->value);
			delete start;
			start=t;
		}
		else{
			printf("%d\n",start->value);
			continue;
		}
		int s=n-1;
		while(s>1){
			if(s==2){
				printf("%d %d\n",start->value,start->next->value);
			}
			int temp=k-1;
			while(temp--){
				start=start->next;
			}
			start->pre->next=start->next;
			start->next->pre=start->pre;
			List *t=start->next;
			printf("%d\n",start->value);
			delete start;
			start=t;
			s--;
		}
		printf("%d\n",start->value);
	}
	return 0;
}

 

 

然後用遞推的方式又寫了一個，把第一個M去掉，並且把所有的編號減去一個M，在最後再加回去

#include <cstdio>
#include <algorithm>
using namespace std;
int main(){
	int i,j;
	int n,k,m;
	while(scanf("%d %d %d",&n,&k,&m)!=EOF){
		if(n==0)
			break;
		int a[2];
		a[0]=0;
		int v=0;
		for(i=2;i<=n-1;i++){
			v=1-v;
			a[v]=(a[1-v]+k)%i;
		}
		printf("%d\n",(a[v]+m)%n+1);
	}
}