poj 2886Who Gets the Most Candies?

题目连接：http://poj.org/problem?id=2886

这道题是模拟约瑟夫环，其具体实现和poj2826差不多的。

我的代码如下：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<memory.h>
int seg_tree[500010<<2];
void build_tree(int l,int r,int id);
void push_tree_up(int id);
void update_tree(int value,int l,int r,int id);
int f[500010];
void get_f();
int num[500010],h;
char str[500010][10];
int cur_n,cur_i,next_i,ans,ans1;
int main()
{
    int n,k,i,j,cur,temp;
    //freopen("d:\\2.txt","r",stdin);
    //freopen("d:\\3.txt","w",stdout);
    memset(f,0,sizeof(f));
    get_f();
    while(scanf("%d%d%*c",&n,&k)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%s%d",str[i],num+i);
        cur_n=n;
        cur_i=n;
        next_i=k;
        build_tree(1,n,1);
        ans=0;
        ans1=1;
        h=1;
        for(i=1;i<=n;i++)
        {
        //  printf("cur_i=%d next_i=%d",cur_i,next_i);
                  if(next_i>=0)
                      cur_i--;
                  cur=(cur_i+next_i%cur_n+cur_n)%cur_n;
              //   printf(" cur=%d\n",cur);
            update_tree(cur+1,1,n,1);
            cur_i=cur;
            h++;


        }
        printf("%s %d\n",str[ans1],ans);

    }

}
void build_tree(int l,int r,int id)
{
    if(l==r)
    {
        seg_tree[id]=1;
        return ;
    }
    int mid=(l+r)>>1;
    build_tree(l,mid,id<<1);
    build_tree(mid+1,r,id<<1|1);
    push_tree_up(id);
}
void push_tree_up(int id)
{
    seg_tree[id]=seg_tree[id<<1]+seg_tree[id<<1|1];
}
void update_tree(int value,int l,int r,int id)
{
    if(l==r)
    {
        seg_tree[id]--;
    // printf("tree[id]=%d",seg_tree[id]);
    //    printf("l=%d\n",l);
            next_i=num[l];
        cur_n--;
        if(ans<f[h])
        {

        ans=f[h];
        ans1=l;
        }
        return ;
    }
    int mid=l+r>>1;
    if(value<=seg_tree[id<<1])
    {
        update_tree(value,l,mid,id<<1);
    }
    else
    {
        update_tree(value-seg_tree[id<<1],mid+1,r,id<<1|1);
    }
    push_tree_up(id);
}
void get_f()
{

    int i,j;
    for(i=1;i<=500000;i++)
    for(j=1;i*j<=500000;j++)
    {
        f[i*j]++;
    }
}