poj 3750 链表

http://poj.org/problem?id=3750小孩报数问题

好不容易写出了单链表版本，提交竟然out time limited。代码如下：

#include <iostream>
#include <string>
using namespace std;

typedef struct child
{
char name[20];
struct child* next;
};
void child_delete(child* p)
{
child* _p = p->next;
p->next = _p->next;
free(_p);
}
int main()
{
int N,_N;
child* head;
child* tail;
head = (child*)malloc(sizeof(child));
tail = head;
tail->next = NULL;

int flag = 1;
int flag2 = 1;
cin>>N;
_N = N;
while(N--)
{
child* stu;
if(flag ==1)
{
stu = head;
flag = 0;
}
else
{
stu = (child*)malloc(sizeof(child));
if(stu == NULL)
return 0;
}
cin>>stu->name;

stu->next = NULL;
if(flag2 == 1)
{
tail = stu;
flag2 = 0;
}
else
{
tail->next = stu;
tail = stu;
}
}
tail->next = head; //形成一个圈
child* pp;
pp = head;
//while(pp)
//{
// cout<<pp->name<<endl;
// pp = pp->next;
//}
int w,s,_s;
scanf("%d,%d",&w,&s);

int _w = 1;
pp = head;
while(_w<w)
{
pp = pp->next;
_w++;
}
_s = s;
while(_N--)
{
s = _s;
s--;
s--;
while(s--)
{
pp = pp->next;
}

cout<<pp->next->name<<endl;

//删除pp指向的下一个指针，并处理尾指针
child_delete(pp);
pp = pp->next;
}
return 0;
}

 于是在网上搜了下解法，原来，不用链表也可以。这种循环控制的想法真好。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
char a[70][100];
int s[70];
int n,i,k;
int x,y;
scanf("%d",&n);
for(i = 1;i<=n;i++)
{
scanf("%s",a[i]);
s[i] = 0;
}
scanf("%d,%d",&x,&y);
k=0;
int count=0;
for(i=x;;i++)
{
if(i>n) i=1;
if(count==n)
break;
if(s[i]==0)
k++;
if(k==y)
{
printf("%s\n",a[i]);
s[i]=1;
count++;
k=0;
}
}
return 0;
}

还算有收获，记录在此。