Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B: Persistent Numbers

	1	2	3	4	5	6	7	8	9
1	1	2	3	4	5	6	7	8	9
2	2	4	6	8	10	12	14	16	18
3	3	6	9	12	15	18	21	24	27
4	4	8	12	16	20	24	28	32	36
5	5	10	15	20	25	30	35	40	45
6	6	12	18	24	30	36	42	48	54
7	7	14	21	28	35	42	49	56	63
8	8	16	24	32	40	48	56	64	72
9	9	18	27	36	45	54	63	72	81

The multiplicative persistence of a number is defined by Neil Sloane (Neil J.A. Sloane in The Persistence of a Number published in Journal of Recreational Mathematics 6, 1973, pp. 97-98., 1973) as the number of steps to reach a one-digit number when repeatedly multiplying the digits. Example:

679 -> 378 -> 168 -> 48 -> 32 -> 6.

That is, the persistence of 679 is 5. The persistence of a single digit number is 0. At the time of this writing it is known that there are numbers with the persistence of 11. It is not known whether there are numbers with the persistence of 12 but it is known that if they exists then the smallest of them would have more than 3000 digits.

The problem that you are to solve here is: what is the smallest number such that the first step of computing its persistence results in the given number?

For each test case there is a single line of input containing a decimal number with up to 1000 digits. A line containing -1 follows the last test case. For each test case you are to output one line containing one integer number satisfying the condition stated above or a statement saying that there is no such number in the format shown below.

Sample input

0
1
4
7
18
49
51
768
-1

Output for sample input

10
11
14
17
29
77
There is no such number.
2688

---

P. Rudnicki

找一個最小的數字其每個位數相乘會等於 n, 個位數字特別討論,
接著從 9 開始試除直到 2, 如果仍然除不等於1則輸出無解

#include <stdio.h>
#include <stdlib.h>
int main() {
    char s[3005];
    int i, j;
    while(gets(s)) {
        if(s[0] == '-') break;
        if(s[1] == '\0') {
            printf("1%s\n", s);
            continue;
        }
        for(i = 0; s[i]; i++)   s[i] -= '0';
        int len = i, idx = 0, st = 0;
        char stk[3005];
        for(i = 9; i >= 2; i--) {
            while(1) {
                int mod = 0;
                for(j = st; j < len; j++) {
                    mod = mod*10 + s[j];
                    mod %= i;
                }
                if(!mod) {
                    stk[idx++] = i+'0';
                    mod = 0;
                    for(j = st; j < len; j++) {
                        mod = mod*10 + s[j];
                        s[j] = mod/i;
                        mod %= i;
                    }
                    while(s[st] == 0)   st++;
                } else  break;
            }
        }
        if(st == len-1 && s[st] == 1) {
            for(i = idx-1; i >= 0; i--)
                putchar(stk[i]);
            puts("");
        } else
            puts("There is no such number.");
    }
    return 0;
}