UVa 11069 - A Graph Problem
传送门
题意
有n数，选出集合满足以下条件：
- 相差2或者3
- 尽可能长
思路
只要加上相差两位和三位的情况即可
$dp[i] = dp[i - 2] + dp[i - 3]$
代码
#include <bits/stdc++.h>
#define LL long long
using namespace std;
LL dp[100];
int main()
{
int n;
dp[0] = 1, dp[1] = 1, dp[2] = 2;
for (int i = 3; i <= 76; i ++)
dp[i] = dp[i - 2] + dp[i - 3];
while (~scanf("%d", &n))
printf("%lld\n", dp[n]);
return 0;
}