题目链接：hdu 4828 Grids
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; }转载于:https://www.cnblogs.com/bhlsheji/p/4011857.html
