cf 335B Stadium and Games

题意：有n个队伍比赛，比赛的规则是，如果人数是奇数，则任意两个队伍都要比直接决出冠军(n*(n-1)/2)，如果人数是偶数的话，任意两个队比赛（n/2），给出你比赛次数，让你求n

解题思路：假设队伍大小为（2^p*x）则用枚举p   二分找x 就能得出答案，不过要注意数据范围！！

解题代码：

// File Name: 325b.c
// Author: darkdream
// Created Time: 2013年07月19日 星期五 16时44分33秒

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<ctype.h>
long long int up(long long int  n)
{
    double t = log(n)/log(2);
    return int(t);
}
long long  Pow(long long p)
{
   long long  sum = 1; 
   for(long long i = 1; i <= p ;i ++)
      sum = sum * 2 ;
   return sum ;
}
int  cmp(const void *a , const void *b)
{
   return *(long long *)a - *(long long *)b;
}
long long ans[100] = {0},pre =0  ;
int  fen(long long n , long long p)
{
   long long  up = 0 ; 
   long long  high; 
   if(p == 1)
   { 
       high = n; 
       if(high >= 2 *1e9)
         high = 2 * 1e9; 
   }
   else
   {
       high = n/(p-1) + 1;
       
       if(high >= 2*1e9)
           high = 2*1e9;
   }
   long long mid =  (up + high)/2;
   //printf("%I64d**
",high);
   while(up <= high)
   {
      mid = (up + high )/2;
      if(mid*(mid-1)/2 + mid*(p-1) > n || mid*(mid-1)/2 +mid*(p-1) < 0  || mid*(mid-1)/2+mid*(p-1) < up*(up-1)/2 + up*(p-1) )
          high = mid-1;
      else if(mid*(mid-1)/2 + mid*(p-1) < n)
          up = mid + 1;
      else 
      {
         if(mid % 2 == 0 )
             break;
         pre++;
         ans[pre] = mid*p;
         return 1; 
      }
   }
   return 0 ;
}
int main(){

   //freopen("/home/plac/problem/input.txt","r",stdin);
   //freopen("/home/plac/problem/output.txt","w",stdout);
   long long int n ; 
   scanf("%I64d",&n);
   long long  p = up(n) + 1;
   int ok = 0 ;
   for(long long  i = 0 ;i <= p;i ++)
   {
       if(fen(n,Pow(i)))
           ok = 1;
   }
   if(!ok)
       printf("-1
");
   else
   {
      //qsort(ans+1,pre,sizeof(long long ),cmp);
      for(int i = 1;i <= pre; i ++)
          if(ans[i] != ans[i-1])
              printf("%I64d
",ans[i]);
   }
return 0 ;
}