【OpenJudge 191】【POJ 1189】钉子和小球

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;
}