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 C: Eventually periodic sequence

Given is a function f: 0..N --> 0..N for a non-negative N and a non-negative integer n ≤ N. One can construct an infinite sequence F = f ¹(n), f ²(n), ... f ^k(n) ... , where f ^k(n) is defined recursively as follows: f ¹(n) = f(n) and f ^k+1(n) = f(f ^k(n)).

It is easy to see that each such sequence F is eventually periodic, that is periodic from some point onwards, e.g 1, 2, 7, 5, 4, 6, 5, 4, 6, 5, 4, 6 ... . Given non-negative integer N ≤ 11000000 , n ≤ N and f, you are to compute the period of sequence F.

Each line of input contains N, n and the a description of f in postfix notation, also known as Reverse Polish Notation (RPN). The operands are either unsigned integer constants or N or the variable x. Only binary operands are allowed: + (addition), * (multiplication) and % (modulo, i.e. remainder of integer division). Operands and operators are separated by whitespace. The operand % occurs exactly once in a function and it is the last (rightmost, or topmost if you wish) operator and its second operand is always N whose value is read from input. The following function:

 
            2 x * 7 + N % 

is the RPN rendition of the more familiar infix (2*x+7)%N. All input lines are shorter than 100 characters. The last line of input has N equal 0 and should not be processed.

For each line of input, output one line with one integer number, the period of F corresponding to the data given in the input line.

Sample input

10 1 x N %
11 1 x x 1 + * N %
1728 1 x x 1 + * x 2 + * N %
1728 1 x x 1 + x 2 + * * N %
100003 1 x x 123 + * x 12345 + * N %
0 0 0 N %

Output for sample input

1
3
6
6
369

---

Piotr Rudnicki (from folklore, sometimes attributted to R. Floyd)

多一個後序處理而已。

#include <stdio.h>
#include <map>
using namespace std;
int main() {
    char s[20], ch;
    int N, n, i, j;
    int p[100][2];
    long long stk[100], a, b;
    while(scanf("%d %d", &N, &n) == 2) {
        if(N == 0 && n == 0)
            break;
        int idx = 0, cnt;
        while(scanf("%s%c", s, &ch) == 2) {
            if(s[0] >= '0' && s[0] <= '9')
                sscanf(s, "%d", &p[idx][0]), p[idx][1] = 1;
            else
                p[idx][0] = s[0], p[idx][1] = 0;
            idx++;
            if(ch == '\n')
                break;
        }
        map<int, int> R[512];
        map<int, int>::iterator it;
        int stkIdx = -1;
        for(cnt = 1; ; cnt++) {
            it = R[n&511].find(n);
            if(it == R[n&511].end()) {
                R[n&511][n] = cnt;
            } else {
                printf("%d\n", cnt-it->second);
                break;
            }
            for(j = 0, stkIdx = -1; j < idx; j++) {
                if(p[j][1] == 1)
                    stk[++stkIdx] = p[j][0];
                else {
                    if(p[j][0] == 'x')
                        stk[++stkIdx] = n;
                    else if(p[j][0] == 'N')
                        stk[++stkIdx] = N;
                    else {
                        b = stk[stkIdx--];
                        a = stk[stkIdx--];
                        if(p[j][0] == '+')
                            stk[++stkIdx] = a+b;
                        else if(p[j][0] == '*')
                            stk[++stkIdx] = a*b;
                        else
                            stk[++stkIdx] = a%b;
                        if(stk[stkIdx] >= N)
                            stk[stkIdx] %= N;
                    }
                }
            }
            n = stk[0];
        }
    }
    return 0;
}