NYOJ 529 月赛水题
2012-04-24 08:19 javaspring 阅读(215) 评论(0) 编辑 收藏 举报这道题可以说是月赛时最水的题了,,没什么意思。题目:
flip
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
-
Give you a non-negative integer x and an operation. The only operation you can do is to reverse one bit in binary form of x
once(i.e 1->0, 0->1).your goal is to turn x into x+1.Calculate the minimum times of operations you need to do.- 输入
-
The first line of the input is an integer T indicates the test cases.
Then follow T lines. Each line is a non-negative integer x as described above, note that 0<=x<10^9. - 输出
- Output the minimum times of operations you need to do to reach the goal.
- 样例输入
-
3 1 2 3
- 样例输出
-
2 1 3
#include <iostream> #include <stdio.h> #include <string.h> int numa[50],numb[50]; void fun(int x,int a[50]){ int k=0; while(x){ a[k++]=x%2; x/=2; } } int main(){ int numcase,n; scanf("%d",&numcase); while(numcase--){ scanf("%d",&n); memset(numa,0,sizeof(numa)); memset(numb,0,sizeof(numb)); fun(n,numa); fun(n+1,numb); int s=0; for(int i=0;i<50;++i) if(numa[i]!=numb[i]) s++; printf("%d\n",s); } return 0; }