Problem 92

A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,

44 → 32 → 13 → 10 → 1 → 1
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

How many starting numbers below ten million will arrive at 89?

C++(Faster):

#include <iostream>
#include <cstring>
#include <stack>using namespace std;const int MAXN = 10000000;int arrive[MAXN], chainno[MAXN];
int no;int nextval(int n)
{int sum, v;sum = 0;while(n) {v = n % 10;sum += v * v;n /= 10;}return sum;
}void makechain(int n)
{no++;while(n != 1 && n != 89) {if(chainno[n] > 0) {n = arrive[chainno[n]];break;}chainno[n] = no;n = nextval(n);}arrive[no] = n;
}int main()
{int n;while(cin >> n && n <= MAXN) {memset(arrive, 0, sizeof(arrive));memset(chainno, 0, sizeof(chainno));chainno[1] = 1;arrive[1] = 1;chainno[89] = 2;arrive[2] = 89;no = 2;for(int i=1; i<n; i++)if(i != 1 && i != 89)makechain(i);int count = 0;for(int i=1; i<n; i++)if(arrive[chainno[i]] == 89)count++;cout << count << endl;}return 0;
}

C++:

#include <iostream>
#include <cstring>
#include <stack>using namespace std;const int MAXN = 10000000;int arrive[MAXN];int nextval(int n)
{int sum, v;sum = 0;while(n) {v = n % 10;sum += v * v;n /= 10;}return sum;
}void makechain(int n)
{stack<int> s;while(n != 1 && n != 89) {if(arrive[n] > 0) {n = arrive[n];break;}s.push(n);n = nextval(n);}while(!s.empty()) {int val = s.top();s.pop();arrive[val] = n;}
}int main()
{int n;while(cin >> n && n <= MAXN) {memset(arrive, 0, sizeof(arrive));arrive[1] = 1;arrive[89] = 89;for(int i=1; i<n; i++)makechain(i);int count = 0;for(int i=1; i<n; i++)if(arrive[i] == 89)count++;cout << count << endl;}return 0;
}