九度 1113 关于二叉树节点的个数问题

#include <stdio.h>#include <math.h>int main(){        int n,m,left,right;        int count;        int deep_n,deep_m,deep_diff;        int i, j;        for( scanf("%d%d",&m,&n); n!=0 && m!=0; scanf("%d%d",&m,&n) )        {                count = 0;                left = right = m;                deep_n = (int)(log(n)/log(2)+1); //节点n的深度                deep_m = (int)(log(m)/log(2)+1); //节点m的深度                deep_diff = deep_n - deep_m;     //两者的深度之差//其实可以将m节点看成一个独立的子树的根节点，就能理解此处所想表达的意思了//深度为deep的二叉树的节点个数为2^deep-1个，这样会好理解一点                count += (int)pow(2,deep_diff) - 1;                  for(i=1;i<=deep_diff;++i)                {                        left = 2*left;                        right = 2*right+1;                }                if(right<=n)                        count += right - left + 1;                else if(left<=n)                        count += n - left + 1;                                                printf("%d\n",count);        }    return 0;}