(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; }
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Answer:
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8581146 |