L2-029 independent happiness

Make a square sum for each digit of a decimal number , It's called an iteration . If a decimal number can be obtained by several iterations 1, It's called happiness number .1 It's a happy number . Besides , for example 19 after 1 Iterations get 82,2 After iterations, we get 68,3 After iterations, we get 100, Finally get 1. be 19 Is the number of happiness . obviously , Iterate over a happiness number to 1 All the numbers in the process are happiness numbers , Their happiness depends on the initial number . for example 82、68、100 Happiness is attached to 19 Of . And one personal independence of conduct Of happiness , It is not attached to any other number in a limited range ; Its independence It is the number of happiness attached to it . If this number is still a prime number , Then its independence doubles . for example 19 In the interval [1, 100] Inside is a unique happiness number , Its independence is  2×4=8.

On the other hand , If one is greater than 1 After several iterations, the number of is in a dead cycle , That number is not happy . for example 29 Iterate to get 85、89、145、42、20、4、16、37、58、89、…… so 89 To 58 Formed a dead cycle , therefore 29 Not happy .

This question asks you to write a program , List all unique happiness numbers and their independence in a given range .

Input format ：

Enter the two endpoints of the closed interval in the first line ：1<A<B≤104.

Output format ：

Give the closed interval in increasing sequence  [A,B]  All the unique happiness numbers and its independence . Each pair of numbers takes up one line , Between the numbers 1 Space separation .

If there is no happiness number in the range , Output in one line  SAD.

sample input 1：

10 40

sample output 1：

19 8
23 6
28 3
31 4
32 3

Be careful ： In the example ,10、13 It's all happiness , But they are attached to other numbers （ Such as 23、31 wait ）, So don't output . Other numbers actually depend on other happiness numbers , But because those numbers are not in the given range [10, 40] Inside , So they are unique happiness numbers in a given range .

sample input 2：

110 120

sample output 2：

SAD

Code length limit

16 KB

The time limit

400 ms

Memory limit

64 MB

Ideas ： use book Array to record the status of each data , If he can push it out 1, But he can also be retired from us book What's in it is -2, If he can't launch -1, We store 0, If you can launch 1 And it cannot be withdrawn within this range , Is the value of his independence , And then use map To store previous values , If it happens again, we will jump out of the cycle , Otherwise, follow his steps , During this period, if the previously identified cannot be exported 1 And back to , If you are a maverick at present, mark the previous numbers as -2, Finally, cycle here , Greater than 0 What you want is what you want .

Code ：

#include <bits/stdc++.h>
using namespace std;
int book[10010];
bool sushu(int n){
	if(n==1)return false;
	if(n==2)return true;
	int i;
	for(i=2;i<=sqrt(n);i++){
		if(n%i==0)return false;
	}
	return true;
}
int u;
void judge(int n){
	if(book[n]==0)return ;
	if(book[n]==-2)return ;
	unordered_map<int,int>mymap;
	while(1){
		int s=0;
		while(n){
			int k=n%10;
			s+=k*k;
			n/=10;
		}
		if(book[s]==0)return ;
		if(s==1){
			unordered_map<int,int>::iterator it;
			for(it=mymap.begin();it!=mymap.end();it++){
				book[it->first]=-2;
			}
			book[u]=mymap.size()+1;// Count yourself 
			if(sushu(u))book[u]*=2;
			return ;
		}
		if(mymap.count(s)){book[s]=0;return ;}
		else {
			mymap[s]=1;
			n=s;
		}
	}
}
int main(){
	int l,r;
	fill(book,book+10010,-1);
	scanf("%d%d",&l,&r);
	int i;
    bool flag=0;
	for(i=l;i<=r;i++){
		u=i;
		judge(i);
	}
	for(i=l;i<=r;i++){
		if(book[i]>0){
            flag=1;
			printf("%d %d\n",i,book[i]);
		}
	}
    if(flag==0)printf("SAD");
	system("pause");
}