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]++;}
}