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
GNU = GNU'sNotUnix
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds

Let us define GNU, the recursive acronym for GNU's Not Unix with the following recursive rules: G -> GNU's N -> Not U -> Unix In each step we apply all the rules simultaneously. If a character in a string does not have a rule associated with it (there will be at most one rule per character), it remains in the string.

For example if we start with GNU, we get:

Step	String
0	GNU
1	GNU'sNotUnix
2	GNU'sNotUnix'sNototUnixnix
3	GNU'sNotUnix'sNototUnixnix'sNotototUnixnixnix
4	GNU'sNotUnix'sNototUnixnix'sNotototUnixnixnix'sNototototUnixnixnixnix
...	...

As you can see, the strings are growing larger in every steps. And in certain cases the growth can be quite fast. Fear not, for we're not interested in the entire string. We just want to know the frequency of a particular character after a finite number of steps.

Input

The first line of the input starts with an integer, T(1<=T<=10), the number of test cases. Then T test cases will follow. The first line of each test case will give you R, (1<=R<=10) the number of rules. In each of the next R lines there will be one rule. The rules are written in the following format: "x->S" (without the quotes). Here x is a single character and S is a sequence of characters. You can assume that the ASCII value of the characters lie in the range 33 to 126, that S will contain no more than 100 characters. You can also assume that the symbols '-' and '>' won't appear in S. The rules can be immediately recursive, recursive in a cycle or not recursive at all. After the rules, you'll find an integer Q (1<=Q<=10), the number of queries to follow.. Each of the Q queries will be presented in the format "initial_string x n" in a single line, where initial_string is the string that you have in step 0 (with length between 1 and 100), x is the character whose frequency you should count, and n (1<=n<=10000) is the number of times the rules should be applied. All characters in the initial string will have ASCII values in the range 33 to 126.

Output

For each query, print the number of occurances of the particular character in the result string after applying the rules n times, starting with the initial string. The output will always fit in a 64-bit unsigned integer.

Sample Input	Sample Output
2 3 G->GNU's N->Not U->Unix 2 GNU t 3 GNU N 3 1 A->BAcX 1 ABCcXA c 10000	6 4 20001

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Problem setter: Monirul Hasan, Member of Elite Problemsetters' Panel
Special thanks to: Jimmy Mårdell (Alternate Solution)

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紀錄個數即可，然後轉換的時候也只記錄字母出現的次數。
記得有些字母沒有轉換，是保留的。

#include <stdio.h>
#include <map>
using namespace std;

int main() {
    int t, n, m, i, j;
    char c, s[1024];
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        while(getchar() != '\n');
        map<int, int> g[128];
        while(n--) {
            scanf("%c->%s", &c, s);
            while(getchar() != '\n');
            for(i = 0; s[i]; i++)
                g[c][s[i]]++;
        }
        scanf("%d", &m);
        while(m--) {
            scanf("%s %c %d", s, &c, &n);
            unsigned long long cnt[128] = {}, tmp[128] = {}, num, flag;
            for(i = 0; s[i]; i++)
                cnt[s[i]]++;
            while(n--) {
                for(i = 33; i <= 126; i++) {
                    num = cnt[i], flag = 1;
                    for(map<int, int>::iterator it = g[i].begin();
                        it != g[i].end(); it++) {
                        tmp[it->first] += num * it->second, flag = 0;
                    }
                    tmp[i] += num * flag;
                }
                for(i = 33; i <= 126; i++)
                    cnt[i] = tmp[i], tmp[i] = 0;
            }
            printf("%llu\n", cnt[c]);
        }
    }
    return 0;
}