（Problem 74）Digit factorial chains
The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145
Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:
169 363601 1454 169 871 45361 871 872 45362 872
It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,
69 363600 1454 169 363601 ( 1454) 78 45360 871 45361 ( 871) 540 145 ( 145)
Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
题目大意：
数字145有一个著名的性质：其所有位上数字的阶乘和等于它本身。
1! + 4! + 5! = 1 + 24 + 120 = 145
169不像145那么有名，但是169可以产生最长的能够连接回它自己的数字链。事实证明一共有三条这样的链：
169 363601 1454 169 871 45361 871 872 45362 872
不难证明每一个数字最终都将陷入一个循环。例如：
69 363600 1454 169 363601 ( 1454) 78 45360 871 45361 ( 871) 540 145 ( 145)
从69开始可以产生一条有5个不重复元素的链，但是以一百万以下的数开始，能够产生的最长的不重复链包含60个项。
一共有多少条以一百万以下的数开始的链包含60个不重复项？
//（Problem 74）Digit factorial chains // Completed on Tue, 18 Feb 2014, 04:21 // Language: C11 // // 版权所有（C）acutus (mail: [email protected]) // 博客地址：http://www.cnblogs.com/acutus/ #include<stdio.h> #include<math.h> #include<stdbool.h> #define N 1000000 long long fac[10]; //保存1~ 9阶乘的数组 long long factorial(int n) //计算阶乘函数 { if(n == 1 || n == 0) return 1; else return n * factorial(n - 1); } void init() //初始化数组 { int i; for(i = 0; i <= 9; i++) { fac[i] = factorial(i); } } long long sum(long long n) //计算整数n各位的阶乘的和 { int ans = 0; while(n) { ans += fac[n % 10]; n /= 10; } return ans; } bool fun(int n) { int i, count, t; bool flag = false; count = 0; while(1) { switch(n) { case 1: count += 1; flag = true; break; case 2: count += 1; flag = true; break; case 169: count += 3; flag = true; break; case 1454: count += 3; flag = true; break; case 871: count += 2; flag = true; break; case 872: count += 2; flag = true; break; case 145: count += 1; flag = true; break; default: t = sum(n); if( n == t) { flag = true; break; } else{ n = t; count++; break; } } if(flag) break; } if(count == 60) return true; else return false; } void solve() { int i, count; count = 0; for(i = 2; i <= N; i++) { if(fun(i)) count++; } printf("%d\n", count); } int main() { init(); solve(); return 0; }
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Answer:
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402 |