POJ 1142 Smith Numbers（史密斯数）
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Description |
题目描述 |
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While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University, noticed that the telephone number of his brother-in-law H. Smith had the following peculiar property: The sum of the digits of that number was equal to the sum of the digits of the prime factors of that number. Got it? Smith's telephone number was 493-7775. This number can be written as the product of its prime factors in the following way:
4937775= 3*5*5*65837
The sum of all digits of the telephone number is 4+9+3+7+7+7+5= 42,and the sum of the digits of its prime factors is equally 3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that he named this kind of numbers after his brother-in-law: Smith numbers.
As this observation is also true for every prime number, Wilansky decided later that a (simple and unsophisticated) prime number is not worth being a Smith number, so he excluded them from the definition.
Wilansky published an article about Smith numbers in the Two Year College Mathematics Journal and was able to present a whole collection of different Smith numbers: For example, 9985 is a Smith number and so is 6036.However, Wilansky was not able to find a Smith number that was larger than the telephone number of his brother-in-law. It is your task to find Smith numbers that are larger than 4937775! |
阿尔伯特·威兰斯基是一位理海大学的数学家，在1982年浏览他自己的电话薄时，注意到他的表兄弟（Harold Smith）H. Smith的电话号码有有如下特点：各位上的数字相加等于分解质因数后各位上的数字相加。懂否？史密斯的电话号码是493-7775。这个数字可被分解质因数致如下形式：
4937775= 3*5*5*65837
这个电话号码各位数字的和是4+9+3+7+7+7+5= 42，并且与分解质因数后各位数字的和相等3+5+5+6+5+8+3+7=42。威兰斯基感觉很神奇就以他的表兄弟命名：史密斯数。
不过这个性质对每个质数都成立，因此威兰斯基后来把质数（分解不能）从史密斯数的定义中剔除了。
威兰斯基在the Two Year College Mathematics Journal发表了关于史密斯数的论文并且列出了一整套史密斯数：举个栗子，9985是史密斯数，6036也是。但是威兰斯基没能找到比他表兄弟电话号码4937775更大的史密斯数，你可以当条红领巾！ |
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Input |
输入 |
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The input file consists of a sequence of positive integers, one integer per line. Each integer will have at most 8 digits. The input is terminated by a line containing the number 0. |
输入文件由一列正整数组成，每行一个整数。每个整数最多8位。数字0表示输入结束。 |
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Output |
输出 |
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For every number n > 0 in the input, you are to compute the smallest Smith number which is larger than n, and print it on a line by itself. You can assume that such a number exists. |
对于每个n>0的输入，你要算出大于n的最小史密斯数，输出一行。你可以认为结果是存在的。 |
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Sample Input - 输入样例 |
Sample Output - 输出样例 |
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4937774 0 |
4937775 |
【题解】
首先，这道题是水题，不然就会和某个人一样觉得要用Pollard's rho算法……
注意几点就可以了：
①可以暴力。②素数不是史密斯数。③从n+1开始找。
④题目描述和输入输出分开看，并不是要你找4937775后的史密斯数。
【代码 C++】
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 int prime[1230]; 5 void rdy(){ 6 bool temp[10005]; 7 memset(temp, 0, sizeof(temp)); 8 prime[0] = 2; 9 int i = 3, j, pi = 0; 10 for (i = 3; i < 10005; i += 2){ 11 if (temp[i]) continue; 12 else{ 13 for (j = i << 1; j < 10005; j += i) temp[j] = 1; 14 prime[++pi] = i; 15 } 16 } 17 prime[1229] = 9974; 18 } 19 int digitSum(int now){ 20 int sum = 0; 21 while (now) sum += now % 10, now /= 10; 22 return sum; 23 } 24 int find(int now){ 25 int i = 0, ed = sqrtf(now) + 0.5; 26 if (ed > 9973) ed = 9973; 27 for (; prime[i] <= ed; ++i){ 28 if (now%prime[i] == 0) return prime[i]; 29 } 30 return 0; 31 } 32 int change(int now){ 33 int sum = 0, temp, stp = 0; 34 while (now > 1){ 35 temp = find(now); 36 if (temp) sum += digitSum(temp), now /= temp, ++stp; 37 else sum += digitSum(now), now = 0; 38 } 39 if (stp) return sum; 40 return 0; 41 } 42 int main(){ 43 rdy(); 44 int n; 45 while (scanf("%d", &n)){ 46 if (n++){ 47 while (digitSum(n) != change(n)) ++n; 48 printf("%d\n", n); 49 } 50 else break; 51 } 52 return 0; 53 }
POJ 1142
