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 D: Dice from pennies

Colin likes to play role playing games. These games usually involve dice with varying numbers of sides; a 20 sided die is known as a D20 (marked 1-20 inclusive), a regular six sided die a D6, and so on. Unfortunately, Colin has forgotten his bag of dice, and has only a jar full of pennies. Help him make his savings throw! For each required roll, Colin will tip out the pennies and read off whether they are heads or tails. This will serve as the random input to a function that returns a number fitting the range of the die. For example, to simulate a regular die, the pairing is as follows:

 
HHH = 1
HHT = 2
HTH = 3
HTT = 4
THH = 5
THT = 6

and analogously for other dice using the minimum number of pennies to encode all the needed die numbers. See sample input for examples. Note that some outputs from the coins will not match to a die number when read from the first position. In this case, use the next coins in the list to return a result. For example, in case of 6 sided dice, TTTHHH maps to the value 1, not 5 as someone may expect. The input will consist of multiple sets of input, each set consisting of a number of dice to be rolled and a type (e.g., 2D6 means roll 2 regular 6 sided dice), followed by the status of the pennies in the next line with no spaces. The last set will start with 0D0 and this set should not be processed. Output will be a single number for each set, consisting of the sum of the simulated dice. If there are insufficient coins to determine an outcome, return -1. There will be at most 1000 coins.

Sample input

1D6
HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
1D6
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
1D6
HTTHHHHTHTHTTTTHTHTHHHTHTHTHTHTHTTHTHTTTTTTHTHHTT
2D2
THTHTHHHHTHHHHTTTTTHTHTHHTHTHTTHTHTHHTHTHTHTHHHHH
1D20
THHHTHHHHTTHTTTTTTHHHHTHTHHHHHHHHHTTTTHTHHHHHHTTT
0D0

Output for sample input

1
-1
4
3
18

---

題目完全沒看懂，不過還是 AC 了。

看了討論區的測資，輸入的字串應為當做二元串來讀，
而字串拆成好幾段，每一段應該可以被轉換成二進制數字，

在不超過骰子點數的情況下，要找到 n 組符合的骰子點數，並且加總。

字串拆的方式，根據骰子最大點數，以最少的二進制數字表示。

#include <stdio.h> 
#include <string.h>
using namespace std;
int main() {
	int n, m, i, j;
	char s[1024];
	while(scanf("%dD%d", &n, &m) == 2)  {
		if(n == 0 && m == 0)
			break;
		scanf("%s", s);
		int d = 0;
		for(i = 0; i < 30; i++)
			if((m-1)&(1<<i))
				d = i;
		d++;
		int l = strlen(s);
		int sum = 0, cnt = 0;
		for(i = 0; i+d <= l && n; i += d) {
			int val = 0;
			for(j = 0; j < d; j++) {
				val = val << 1;
				if(s[i+j] == 'T')
					val |= 1;
			}
			val++;
			if(val <= m) {
				sum += val;
				cnt++;
				n--;
			}
		}
		if(n == 0)
			printf("%d\n", sum);
		else
			puts("-1");
	}
	return 0;
}
/*
18D19
TTTHHHHTTTHHHHTTHTTHTTTTHTTHTHTTTHTTTHTTTTTHTTHHTTHTTTTHTTTTTTTTTTHHHHHHHHHHHTT
*/