2.3 KiB
2.3 KiB
本题和 96.不同的二叉搜索树 比较像
- 取模这里很容易出错
- 过程中所用到的数值都有可能超过int,所以要改用longlong
#include <iostream>
#include <vector>
using namespace std;
long long mod = 1e9 + 7;
long long dp(int t, vector<long long>& memory) {
if (t % 2 == 0) return 0;
if (t == 1) return 1;
if (memory[t] != -1) return memory[t];
long long result = 0;
// 枚举左右子树节点的数量
for (int i = 1; i < t; i += 2) {
long long leftNum = dp(i, memory); // 左子树节点数量为i
long long rightNum = dp(t - i - 1, memory); // 右子树节点数量为t - i - 1
result += (leftNum * rightNum) % mod; // 注意这里是乘的关系
result %= mod;
}
memory[t] = result;
return result;
}
int main() {
int n;
cin >> n;
vector<long long> memory(n + 1, -1);
cout << dp(n, memory) << endl;
}
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int MOD = 1000000007;
int main() {
int num;
cin >> num;
if (num % 2 == 0) {
cout << 0 << endl;
return 0;
}
vector<long long> dp(num + 1, 0);
dp[1] = 1;
for (int i = 3; i <= num; i += 2) {
for (int j = 1; j <= i - 2; j += 2) {
dp[i] = (dp[i] + dp[j] * dp[i - 1 - j]) % MOD;
}
}
cout << dp[num] << endl;
return 0;
}
第二题的代码
#include <bits/stdc++.h> using namespace std;
long fastexp(long base,long n,long mod){ long answer = 1; while(n > 0){ if(n % 2 == 1){ answer = (answer * base) % mod; } base = (base * base) % mod; n /= 2; } return answer; } int kawaiiStrings(int n) { // write code here std::vector f(n + 1), g(n + 1), h(n + 1); long mod = 1000000007; for (long i = 2; i <= n; i++) g[i] = (g[i - 1] * 2 + (i - 1) * fastexp(2,i-2,mod)) % mod; for (long i = 3; i <= n; i++) f[i] = ((f[i - 1] * 3) % mod + g[i - 1]) % mod; for (long i = 3; i <= n; i++) h[i] = (fastexp(3, i - 3, mod) + h[i - 1] * 3 - h[i - 3]) % mod; return (f[n]-h[n]+mod)%mod;
}
int main(){ int n; cin >> n; cout << kawaiiStrings(n) << endl; return 0; }