poj 2886Who Gets the Most Candies?

http://poj.org/problem?id=2886

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define maxn 500100
 using namespace std;

 typedef long long ll;
 const int prime[]= {,,,,,,,,,,,,,,,};
 int n,k;
 struct node
 {
     int l,r,num;
 } tree[maxn*];

 ll maxsum,bestnum;
 char name[maxn][];
 int a[maxn];

 void getantiprime(ll num, ll k,ll sum,int limit)
 {
      //num：当前枚举到的数，k：枚举到的第k大的质因子；sum：该数的约数个数；limit：质因子个数上限；
     int i;
     ll temp;
     if(sum > maxsum)
     {
         maxsum = sum;
         bestnum = num; //如果约数个数更多，将最优解更新为当前数；
     }
     if(sum==maxsum && bestnum > num)
         bestnum = num; //如果约数个数相同，将最优解更新为较小的数；
     if(k > )
         return;
     temp = num;
     for(i=; i<=limit; i++) //开始枚举每个质因子的个数；
     {
         if(temp*prime[k] > n)
             break;
         temp = temp * prime[k]; //累乘到当前数；
         getantiprime(temp, k+, sum*(i+), i); //继续下一步搜索；
     }
 }

 void build(int i,int l,int r)//建树
 {
     tree[i].num=r-l+;
     tree[i].l=l;
     tree[i].r=r;
     if(l<r)
     {
         int mid=(l+r)/;
         build(i+i,l,mid);
         build(i+i+,mid+,r);
     }
 }

 int search1(int num,int i)//查找原始序号
 {
     tree[i].num--;
     if(tree[i].l==tree[i].r)
     {
         return tree[i].l;
     }
     if(num<=tree[i+i].num)
         return search1(num,i+i);
     return search1(num-tree[i+i].num,i+i+);
 }

 int main()
 {
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         getantiprime(,,,);//找到n以内反素数最大的；
         build(,,n);
         for(int i=; i<=n; i++)
         {
             scanf("%s %d",name[i],&a[i]);
         }
         int id;
         for(int i=; i<bestnum; i++)//约瑟夫原理
         {
             n--;
             id=search1(k,);
             if(n==) break;
             if(a[id]>)
                 k=(k-+a[id]-)%n+;
             else
                 k=((k-+a[id])%n+n)%n+;
         }
         printf("%s %lld\n",name[id],maxsum);
     }
     return ;
 }