（Problem 92）Square digit chains

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?

 

题目大意:

通过将一个数各位的平方不断相加，直到遇到已经出现过的数字，可以形成一个数字链。

例如：

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

因此任何到达1或89的数字链都会陷入无限循环。令人惊奇的是，以任何数字开始，最终都会到达1或89。

以一千万以下的数字n开始，有多少个n会到达89？

 

算法一：常规方法，从2~10000000逐个判断，同时统计结果

#include<stdio.h>   

#define N 10000000

int fun(int n)
{
	int t, sum;
	sum = 0; 
	while(n) {
		t = n % 10;
		sum += t * t;
		n /= 10;
	}
	return sum;
}

void solve(void)
{
	int i, sum, t;
	sum = 0;
	for(i = 2; i < N; i++) {
		t = fun(i);
		while(1) {
			if(t == 89) {
				sum++;
				break;
			} else if(t == 1) {
				break;
			} else {
				t = fun(t);
			}
		}
	}
	printf("%d\n",sum);
}

int main(void)
{
	solve();
	return 0;
}

 

算法二（优化）：使用一个bool型数组，保存每次结果，由于最大的中间数为9999999产生的：9^2*7 = 567，所以bool型数组的大小开到600足够

#include <stdio.h>
#include <stdbool.h>      

#define N 10000000

bool a[600] = {false};

int fun(int n)
{
	int t, sum;
	sum = 0; 
	while(n) {
		t = n % 10;
		sum += t * t;
		n /= 10;
	}
	return sum;
}

void solve(void)
{
	int i, sum, t, temp;
	sum = 0;
	for(i = 2; i < N; i++) {
		t = fun(i);
		temp = t;
		if(a[temp]) {
			sum++;
		} else {
			while(1) {
				t = fun(t);
				if(a[t] || t == 89) {
					a[temp] = true;
					sum++;
					break;
				} else if(t == 1) {
					break;
				} else {
				}
			}
		}
	}
	printf("%d\n",sum);
}

int main(void)
{
	solve();
	return 0;
}

 

Answer:

8581146

Completed on Sun, 24 Nov 2013, 10:26