//给⼀组组数，仅仅有两个数仅仅出现了一次。其它全部数都是成对出现的,找出这两个数。#include 
     
      int find_one_pos(int num)     //找一个为为1的位置{	int n = 0;	while(num)	{		if (num & 1 == 1)			break;		else		{			n++;			num >>= 1;		}	}	return n;}void find_two_differ(int arr[], int len, int *num1, int *num2){	int i = 0;	int pos = 0;	int ret = 0;	*num1 = 0;	*num2 = 0;	for (i = 0; i < len; i++)	{		ret ^= arr[i];	}	pos = find_one_pos(ret);	for (i = 0; i < len; i++)	{		if (arr[i] & (1 << pos))			*num1 ^= arr[i];		else			*num2 ^= arr[i];	}}int main(){	int arr[] = { 1, 2, 3, 4, 5, 1, 2, 3 };	int ret1 ;	int ret2 ;	find_two_differ(arr, sizeof(arr) / sizeof(arr[0]), &ret1, &ret2);	printf("num1=%d\nnum2=%d\n", ret1, ret2);	return 0;}