赛氪OJ-专注于算法竞赛的在线评测系统 (saikr.com)

题解

因为题目要求同一侧不能挨着做，因此我们可以考虑如下：

历程

我刚开始按照题解写了如下两段代码，全部没过：

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10;
const int MOD = 1e9 + 7;
int dp[N];

signed main() {
	int n;
	cin >> n;
	dp[0] = 1;
	dp[1] = 2;

	for (int i = 2; i <= n; i++) {
		dp[i] = dp[i - 1] + dp[i - 2] ;
	}

	cout << dp[n]*dp[n]% MOD ;

	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10;
const int MOD = 1e9 + 7;
int dp[N];

signed main() {
	int n;
	cin >> n;
	dp[0] = 1;
	dp[1] = 2;

	for (int i = 2; i <= n; i++) {
		dp[i] =( dp[i - 1] + dp[i - 2]) % MOD;
	}

	cout << dp[n]*dp[n] ;

	return 0;
}

下面为AC代码：

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10;
const int MOD = 1e9 + 7;
int dp[N];

signed main() {
	int n;
	cin >> n;
	dp[0] = 1;
	dp[1] = 2;

	for (int i = 2; i <= n; i++) {
		dp[i] = (dp[i - 1] + dp[i - 2]) % MOD;
	}

	cout << dp[n]*dp[n]%MOD ;

	return 0;
}