POJ 2886 Who Gets the Most Candies? 线段树

题目： http://poj.org/problem?id=2886

左右转的果断晕，题目不难，关键是准确的转啊转。因为题目要求输出约数个数最多的数，所以预处理[1,500000]的约数的个数就行了。还有利用反素数的做法，太专业了，还是暴力预处理吧。。。

 #include <stdio.h>
 #include <string.h>
 #include <math.h>

 const int MAXN = ;

 struct Tree_Node
 {
     int left, right, sum;
     Tree_Node(){}
     Tree_Node(int left, int right, int sum)
     {
         this->left = left;
         this->right = right;
         this->sum = sum;
     }
 }tree[MAXN<<];

 void build(int left, int right, int step)
 {
     tree[step] = Tree_Node(left, right, right-left+);
     if(tree[step].left == tree[step].right)
         return;
     int mid = left + (right-left) / ;
     build(left, mid, step<<);
     build(mid+, right, step<<|);
 }

 int query(int num, int step)
 {
     tree[step].sum--;
     if(tree[step].left == tree[step].right)
     {
         return tree[step].left;
     }
     if(num <= tree[step<<].sum)
         return query(num, step<<);
     else
         return query(num - tree[step<<].sum, step<<|);
 }

 int div[MAXN];
 void cnt_divisible()
 {
     int x = sqrt(MAXN);
     for(int i = ; i <= x; i++)
     {
         int y = MAXN / i;
         for(int j = i+; j <= y; j++)
             div[i*j] += ;
         div[i*i]++;
     }
 }

 char name[MAXN][];
 int val[MAXN];
 int main()
 {
     cnt_divisible();
     int n, k;
     while(scanf("%d %d", &n, &k) != EOF)
     {
         build(, n, );
         for(int i = ; i <= n; i++)
             scanf("%s %d", name[i], &val[i]);
         int ans = , num = , cnt = ;
         while(n > )
         {
             n--;
             int x = query(k, );
             if(div[++cnt] > ans)
             {
                 ans = div[cnt];
                 num = x;
             }
             if(n == )
                 break;
             if(val[x] > )
                 k = (k + val[x] - ) % n + ;
             else
                 k = ((k + val[x] - ) % n + n) % n + ;
         }
         printf("%s %d\n", name[num], ans);
     }
     return ;
 }