HDOJ2041 超级楼梯(dp & 打表)

超级楼梯

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38263    Accepted Submission(s): 19688

Problem Description

有一楼梯共M级，刚开始时你在第一级，若每次只能跨上一级或二级，要走上第M级，共有多少种走法？

 

Input

输入数据首先包含一个整数N，表示测试实例的个数，然后是N行数据，每行包含一个整数M（1<=M<=40）,表示楼梯的级数。

 

Output

对于每个测试实例，请输出不同走法的数量

 

Sample Input

2
2
3

 

Sample Output

1
2

 

其实就是个斐波那契数列，我直接打表了＝ ＝

AC代码（打表）：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
using namespace std;
typedef long long ll;
int n;
ll dp[] = {0, 0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155};
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	while(t--) {
		scanf("%d", &n);
		printf("%lld\n", dp[n]);
	}
	return 0;
}

AC代码（dp）：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
using namespace std;
typedef long long ll;
int n;
ll dp[45];
int main(int argc, char const *argv[])
{
	dp[2] = 1, dp[3] = 2;
	for(int i = 4; i <= 41; ++i)
		dp[i] = dp[i - 1] + dp[i - 2];
	int t;
	scanf("%d", &t);
	while(t--) {
		scanf("%d", &n);
		printf("%lld\n", dp[n]);
	}
	return 0;
}