絎?姝?鎶婁笂闈㈢殑n-1涓洏縐誨埌絎?鍙鋒潌涓?
絎?姝?鎶婄n涓洏浠?縐誨埌2:
鎴戜滑璁緁(n)涓烘妸n涓洏浠?縐誨埌3鎵闇瑕佺殑姝ユ暟錛屽綋鐒朵篃絳変簬浠?縐誨埌1鐨勬鏁般?br>
鐢變笂闈㈢殑鍥炬垜浠彲浠ョ湅鍒?瑕佹兂鎶婄n涓洏浠?縐誨埌3錛岄渶瑕?涓楠?:
1.) 鎶婂墠n-1涓粠1縐誨姩3 .
2.) 絎琻涓洏瑕佷粠1->2->3緇忓巻2姝?
3.) 鑰屽墠n-1涓洏闇瑕佸厛 3->1 ( 榪欐槸涓轟簡緇?絎琻涓洏璁╄礬 ), 鏈鍚庡啀 1->3銆?br>
∴f(n) = 3 × f(n-1) + 2;
f(1) = 2;
f(n) = 3 × f(n-1) + 2
f(1) = 2
=>
f(n) + 1 = 3 × [f(n-1) + 1]
f(1) + 1 = 2 + 1 = 3
=>
f(n) + 1 = 3n
=>
f(n) = 3n - 1
鏈鍚庤創涓婁唬鐮?:
//MiYu鍘熷垱, 杞笘璇鋒敞鏄?: 杞澆鑷?______________鐧界櫧銇眿
#include <iostream>
#include <cmath>
using namespace std;
long long myPow ( int n , int e )
{
long long mlt = 1;
for ( int i = 1; i <= e ; ++ i )
{
mlt *= n;
}
return mlt;
}
int main ()
{
int N;
while ( cin >> N )
{
cout << myPow ( 3, N ) - 1 << endl;
}
return 0;
}