HDU2047 阿牛的EOF牛肉串

题目:https://blog.csdn.net/qq_40932661?t=1

表面上看去似乎无从下手。但是可以从前面地推出后面的

递推:

假如涂第N个位置,有两种可能,①涂O ②不涂O。

如果涂O的话,前面不能是O,只能是E或F两种, 即2*f(n-2)

不涂O的话,N位置可以放E或F两种即2*f(n-1)

所以 f[n] = 2*(f[n-1] + f[n-2])

!!!输入挂是一个坑点。刚入门。才知道while(scan_d(n))   不能是while((~scan_d)),  否则OLE

AC代码:

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define FO freopen("in.txt", "r", stdin);
#define lowbit(x) (x&-x)
typedef vector<int> VI;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const ll mod=1000000007;
const int inf = 0x3f3f3f3f;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
template <class T>
inline bool scan_d(T &ret)
{
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0; //EOF
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
inline void out(ll x)
{
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}

long long f[50];
int main() {
	f[1] = 3;
	f[2] = 8;
	rep(i, 3, 45) {
		f[i] = 2*(f[i-1] +f[i-2]);
	}
	int n;
	while(scan_d(n)) {
		out(f[n]);
		printf("\n");
	}
} 


 

posted @ 2018-07-28 10:31  Frontierone  阅读(100)  评论(0编辑  收藏  举报