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; }