Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8440 | Accepted: 5093 |
Description
Input
Output
Sample Input
2 8 10
Sample Output
1 15 9 11
==================================================================
这题就是Lowbit函数的使用,理解了Lowbit,这题就很容易了
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #define MAxx 1000 int Lowbit(int x) { return x&(-x); } int main() { int i,j,n,m; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&m); int cas=Lowbit(m); printf("%d %d\n",m-cas+1,m+cas-1); } return 0; }
poj 2309 BST(数学题),布布扣,bubuko.com
原文地址:http://www.cnblogs.com/ccccnzb/p/3837742.html