【OpenJudge 191】【POJ 1189】钉子和小球

时间：2021-08-19 17:38:03

http://noi.openjudge.cn/ch0405/191/

http://poj.org/problem?id=1189

一开始忘了\(2^{50}\)没超long long差点写高精度QvQ

很基础的dp，我先假设有\(2^n\)个球，分开时就分一半，这样每次都能除开。

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 53;

int in() {

	int k = 0, fh = 1; char c = getchar();

	for(; c < '0' || c > '9'; c = getchar())

		if (c == '-') fh = -1;

	for(; c >= '0' && c <= '9'; c = getchar())

		k = k * 10 + c - 48;

	return k * fh;

}

char c[N][N];

ll f[N][N];

int n, m;

int main() {

	n = in(); m = in();

	char ch;

	for(int i = 1; i <= n; ++i)

		for(int j = 1; j <= i; ++j) {

			for(ch = getchar(); ch != '*' && ch != '.'; ch = getchar());

			c[i][j] = ch;

		}

	f[1][1] = 1ll << n;

	for(int i = 2; i <= n + 1; ++i)

		for(int j = 1; j <= i; ++j) {

			if (j - 1 > 0 && c[i - 1][j - 1] == '*')

				f[i][j] += f[i - 1][j - 1] >> 1;

			if (j < i && c[i - 1][j] == '*')

				f[i][j] += f[i - 1][j] >> 1;

			if (i > 2 && 1 <= j - 1 && j - 1 <= i - 2 && c[i - 2][j - 1] == '.')

				f[i][j] += f[i - 2][j - 1];

		}

	ll fz = f[n + 1][m + 1], fm = 1ll << n;

	while (fz % 2 == 0 && fm % 2 == 0)

		fz >>= 1, fm >>= 1;

	if (fz == 0) fm = 1;

	printf("%I64d/%I64d\n", fz, fm);

	return 0;

}

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