输入:整数A
输出:整数B
条件:A和B的二进制1的个数相同,且A和B之间的距离|A-B|最小。
#include <iostream>#include <math.h>using namespace std;int sameSumOfOne(int a){if(a==0)return 0;int b=0;int pos=0;int bit_num=sizeof(int)*8;if((a&1)==0){while((a&(1<<pos))==0 &&(pos<bit_num))pos++;//cout<<"0_pos="<<pos<<endl;if(po***it_num)return a;if(po***it_num-1)return 1;b=a-(1<<pos)+(1<<(pos-1));}else{while((a&(1<<pos))&&(pos<bit_num))pos++;//cout<<"1_pos="<<pos<<endl;// all bit is 1if(po***it_num)return a;if(po***it_num-1)return -2;b=a-(1<<(pos-1))+(1<<pos);}return b;}int main(){int maximum=(int)((unsigned int)-1 >> 1U);int minimum=(int)~((unsigned int)-1 >> 1U);int a[]={0,1,2,3,4,5,6,7,8,9,10,maximum,minimum,-1,-2};int n=sizeof(a)/sizeof(a[0]);for(int i=0;i<n;i++)cout<<a[i]<<" "<<sameSumOfOne(a[i])<<endl;cout<<endl;return 0;}
public class SameNumsOf1{ public static int getSameNumsOf1(int a){ if(a==0) return 0; int location = 0; int b; while(((a>>location)&1)==0) location++; if(location>0){ b = a-(1<<location)+(1<<--location); }else{ while(((a>>location)&1)==1) location++; b = a+1+(1<<--location)-1; } return b; } public static void main(String[] args){ int[] array = {0,1,2,3,4,5,6,7,8,9,10,11,12}; for(int num:array){ System.out.printf("(%d,%d) ",num,getSameNumsOf1(num)); } } }