hdu 4828 Grids(拓展欧几里得+卡特兰数)
题目大意:略。
解题思路:将上一行看成是入栈,下一行看成是出栈,那么执着的方案就是卡特兰数,用递推的方式求解。
#include <cstdio>
#include <cstring>
typedef long long ll;
const int N = 1000005;
const ll MOD = 1e9+7;
ll dp[N];
ll extendGcd(ll a, ll b, ll& x, ll& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
ll d = extendGcd(b, a%b, y, x);
y -= a / b * x;
return d;
}
ll solve (ll n) {
ll x, y;
ll tmp = extendGcd(n + 1, MOD, x, y);
x = (x % MOD + MOD) % MOD;
return x;
}
void init () {
dp[1] = 1;
dp[2] = 2;
for (ll i = 3; i < N; i++)
dp[i] = (dp[i-1] * (4 * i - 2) % MOD * solve(i)) % MOD;
}
int main () {
int cas, n;
scanf("%d", &cas);
init();
for (int i = 1; i <= cas; i++) {
scanf("%d", &n);
printf("Case #%d:\n%lld\n", i, dp[n]);
}
return 0;
}