POJ 3517 And Then There Was One（约瑟夫环-递推or模拟）

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POJ 3517

题目： n k m

数字1到n成环，先叉数字m，往下数k个，直到最后只有一个数字，输出它。

链表模拟：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#define MAXN 200001#define MOD  1000000007#define INF  0x7fffffff#define EPS  1e-8#define PI acos(-1.0)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define bug(a) cout<<"bug---->"<<a<<endl;#define FIN             freopen("datain.txt","r",stdin);#define FOUT            freopen("dataout.txt","w",stdout);#define mem(a,b)        memset(a,b,sizeof(a))//#pragma comment         (linker,"/STACK:102400000,102400000")typedef long long LL;typedef unsigned long long ULL;using namespace std;struct Link{    int data;    Link* next;    Link* pre;}node[10001];int main(){    int n,k,m;    while(scanf("%d%d%d",&n,&k,&m),n||k||m)    {        for(int i=1;i<=n;i++)                            //构建双向循环链表        {            node[i].data=i;            node[i].next=(i==n)?&node[1]:&node[i+1];            node[i].pre=(i==1)?&node[n]:&node[i-1];        }        Link* p=&node[m];        p->pre->next=p->next;        p->next->pre=p->pre;        p=p->next;        int loop=k;        int t=1;        while(p->next!=p)        {            if(k%(n-t)==0)            //优化，若无会TLE                loop=k;            else                loop=k%(n-t);            for(int i=1;i<loop;i++)                p=p->next;            p->pre->next=p->next;            p->next->pre=p->pre;            p=p->next;            t++;        }        printf("%d\n",p->data);    }    return 0;}

递推：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<string>#include<queue>#define N#define INF#define ll __int64#define eps 1e-12#define PI 4.0*atan(1.0)using namespace std;int main(){    int n,m,k;    while(~scanf("%d%d%d",&n,&k,&m))    {        if(n==0 && m==0 && k==0)            break;        int s=0;        for(int i=2;i<=n-1;i++)            s=(s+k)%i;        printf("%d\n",(s+m)%n+1);    }    return 0;}

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