2013-02-19 20:07:37Morris
[UVA][計數] 10625 - GNU = GNU'sNotUnix
Problem
C
GNU = GNU'sNotUnix
Input: Standard Input
Output: Standard
Output
Time Limit: 2 Seconds
![alternative image of the Head of a GNU [http://www.gnu.org/graphics/gnu-alternative.jpg]](http://uva.onlinejudge.org/external/106/p10625.jpg) |
Let us define GNU, the recursive acronym for GNU's Not Unix with the
following recursive rules:
|
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 |
Problem setter: Monirul Hasan, Member of Elite Problemsetters' Panel
Special thanks to: Jimmy Mårdell (Alternate Solution)
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;
}