九度 1389

这是一个变形题；

设f[n]为跳到n阶台阶的方法，另f[0] = 1;

当n = 1时，f[n] = f[1] = f[0] = 1;

当n = 2时，f[n] = f[2] = f[0] + f[1] = 2;

当n = 3时，f[n] = f[3] = f[0] + f[1] + f[2] = 4;

······························

当n = k时，f[n] = f[k] = f[k-1] + f[k-2] + ·················f[1] + f[0];

因此f[n] = f[n-1] + f[n-2] +····················f[0]-------------------(1)

而f[n-1] = f[n-2] + f[n-3] + ····················f[0]------------------(2)

(1)式减去(2)式：

f[n] = f[n-1] + f[n-1] = 2 * f[n-1] = 2 * 2  * f[n-2] = 2 * 2 * 2 * f[n-3] =·····2^(n-1) * f[1]，而f[1] = 1;

因此：f[n] = 2^(n-1) = (1LL << (n-1));

注意：只能用long long 类型，_in64d，不行！输出用"%lld"!

故有下面两个代码：

代码一：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAXN = 55;long long f[MAXN];void Pre_Slove(){    int i;    memset(f, 0, sizeof(f));    f[0] = 1;    f[1] = 1;    for (i = 2; i <= 55; ++i)        f[i] = 2 * f[i-1];}int main(){int n;Pre_Slove();while (~scanf("%d", &n)){printf("%lld\n", f[n]);}return 0;}

代码二：

#include <stdio.h>int main(){int n;while (scanf("%d", &n) != EOF){    printf("%lld\n", 1LL << (n-1));}return 0;}