递推+高精度+找规律 UVA 10254 The Priest Mathematician

题目传送门

 /*
     题意：汉诺塔问题变形，多了第四个盘子可以放前k个塔，然后n-k个是经典的汉诺塔问题，问最少操作次数
     递推+高精度+找规律：f[k]表示前k放在第四个盘子，g[n-k]表示经典三个盘子，2 ^ (n - k) - 1
             所以f[n] = min (f[k] * 2 + g[n-k])，n<=10000，所要要用高精度，另外打表能看出规律
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-18 9:14:21
 * File Name     :UVA_10254.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;

 struct bign {
     short s[MAXN] , len ;
     bign () { memset ( s , , sizeof ( s ) ) ; len =  ; }
     bign operator = (const char *num) {
         len = strlen ( num ) ;
         for ( int i =  ; i < len ; i ++ ) s[i] = num[len-i-] - '' ;
         return *this ;
     }
     bign operator = (int num) {
         char s[MAXN];
         sprintf (s , "%d" , num);
         *this = s ;
         return *this ;
     }
     bign(const char *num) { *this = num ; }
     bign(int num) { *this = num ; }
     string str () const {
         string res ;
         res = "" ;
         for (int i = ; i < len; i ++) res = (char) (s[i] + '') + res ;
         if (res == "") res = '';
         return res ;
     }
     bign operator + (const bign& b) const {
         bign c ;
         c.len =  ;
         for(int i = , g = ; g || i < max (len, b.len); i ++) {
             int x = g ;
             if (i < len) x += s[i] ;
             if (i < b.len) x += b.s[i] ;
             c.s[c.len++] = x %  ;
             g = x /  ;
         }
         return c ;
     }
     void print() {
         for(int i = len - ; i >= ; i --) printf("%hd", s[i]);
         printf("\n");
     }
 }f[];

 int main(void)  {       //UVA 10254 The Priest Mathematician
     bign g = ; f[] = ;
     for (int i=, j=; i<=; j++, g=g+g)    {
         for (int k=; k<=j && i<=; k++,i++)    {
             f[i] = f[i-] + g;
         }
     }
     int n;
     while (scanf ("%d", &n) == )   {
         f[n].print ();
     }

     return ;
 }