Example 1:
Input:
2
3
7
Output:
1 3
2 3 6
Description: . 3 =. 3, (. 3 |. 6). 7 =
Meaning of the questions: digital output as little as possible, according to their position or results of digital input a.
:( solution to a problem that can not read explanations, see Shandong Gangster code to read it) First of all: There are two strange phenomenon, is a binary digit (my definition is the first 0, that is an even number) and only 2 1, and respectively at the even bit and odd bit, this number must be a multiple of 3. If a number a (a> 3) at the even bit and odd bit are 1, and the number 1 is greater than or equal to 1, as a% 3 == 1, only a take on even bit 1 to 3% == 0; when a% 3 == 2, took only 1% to 3 == 0 on an odd bit.
1, if the input is a multiple of 3, then a direct output;
2, if the input is not a multiple of three, the number of recording a binary number and position on the odd bit and the even bit is 1, then the following actions:
If the odd bit and the even bit 1 are present, the builder num1 = 0, num2 = a, and if a% 3 == 1, then on a 1 num2 away even bit, odd bit and num1 to respective even bit an existing 1; if a% 3 == 2, will be a 1 on the odd bit away num2, num1 to each even bit and odd bit an existing one;
If there is only one position on the odd or even bit, similarly configured num1 = 0, num2 = a, and progressively take 1 to existing num1 num2 1 on, until num2% 3 == 0, num2 stop away 1 , continued existing num1 1 until the end of num1% 3 = 0, complete the construction.
AC Code:
1 #include<bits/stdc++.h> 2 #define LL long long 3 #define pb push_back 4 using namespace std; 5 LL num1,num2; 6 void solve(LL x) 7 { 8 num1=0,num2=x; 9 vector<int>odd,even; 10 for(int i=0;i<=63;i++){//将奇数位和偶数位上的1的位置取出并保存 11 if((x>>i)&1){ 12 if(i%2)odd.pb(i); 13 else even.pb(i); 14 } 15 } 16 if(odd.size()&&even.size()){//奇数位和偶数位上都存在1的情况 17 if(x%3==1){ 18 num2^=(1LL<<even[0]); 19 num1^=(1LL<<even[0]); 20 num1^=(1LL<<odd[0]); 21 } 22 else{ 23 num2^=(1LL<<odd[0]); 24 num1^=(1LL<<even[0]); 25 num1^=(1LL<<odd[0]); 26 } 27 } 28 else if(odd.size()){//只有奇数位上存在1的情况 29 int i; 30 for(i=0;i<odd.size();i++){ 31 num1^=(1LL<<odd[i]),num2^=(1LL<<odd[i]); 32 if(num2%3==0){ 33 i++;break; 34 } 35 } 36 for(;i<odd.size();i++){ 37 if(num1%3==0)break; 38 num1^=(1LL<<odd[i]); 39 } 40 } 41 else{//只有偶数位上存在1的情况 42 int i; 43 for(i=0;i<even.size();i++){ 44 num1^=(1LL<<even[i]),num2^=(1LL<<even[i]); 45 if(num2%3==0){ 46 i++;break; 47 } 48 } 49 for(;i<even.size();i++){ 50 if(num1%3==0)break; 51 num1^=(1LL<<even[i]); 52 } 53 } 54 } 55 int main() 56 { 57 LL t,n; 58 scanf("%lld",&t); 59 while(t--){ 60 scanf("%lld",&n); 61 vector<LL>num; 62 if(n%3==0){ 63 num.pb(n); 64 } 65 else { 66 solve(n);//a%3!=0的情况下进行构造。 67 num.pb(num1);num.pb(num2); 68 } 69 if(num.size()==1) 70 printf("1 %lld\n",num[0]); 71 else if(num.size()==2) 72 printf("2 %lld %lld\n",num[0],num[1]); 73 } 74 return 0; 75 }