题意:。。。
析:由于能最后一个是特殊的,所以前n-1个都是不变的,只是减少了最后一个盘子的次数,所以根据上一个题的结论 答案就是dp[n-1] + 2。
上一题链接:http://www.cnblogs.com/dwtfukgv/p/6283879.html
代码如下;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <unordered_map> #include <unordered_set> #define debug() puts("++++"); #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 40 + 5; const int mod = 2000; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL dp[maxn]; void init(){ dp[1] = 2; for(int i = 2; i < 21; ++i) dp[i] = dp[i-1] * 3LL + 2LL; } int main(){ init(); int T; cin >> T; while(T--){ cin >> n; cout << dp[n-1] + 2LL << endl; } return 0; }