[2016-03-03] [UVA] [1374] [Power calculus]

[2016-03-03] [UVA] [1374] [Power calculus]

• Time: 2016-03-03-16:14:01 Thursday
  
• Topic Number: UVA 1374
  
• The main topic: Give x exponent N, ask, x after how many times to get x^n
  
• Input: N
  
• Output: Number of times
  
• Analysis:
• Find all possible ways to multiply, use DFS, proper pruning optimization, and become ida*
• The multiplication of x becomes the addition of the exponent
• Each time the square, the minimum number of times is the number of squares. This for Maxd beginnings
• Method:
• Enumerates the number of occurrences of each bit, the current word count, plus the number of times enumerated, into the next layer of Dfs
• Pruning: return 0 if expected minimum number of times plus current count > Maxd
• Problems encountered during the problem solving process:
  
• There is no need to take the maximum value to determine the remaining, the direct current value is continuously squared (exponential addition) 1<< (maxd-cur-1) times can be

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> using namespace std; typedef Long long LL; #define The CLR (x , Y) memset ((×), (y),sizeof((x))) #define for (x, Y, z) for ( int.) = (y);(x) < (*); + + (x)) #define FORD ( x, Y, z) for(int.) = (y);(x) >= (z);--(x)) #define FOR2 ( x, Y, z) for((×) = (y);(x) < (z); + + (x) )#define FORD2 ( x, Y, z) for((>=) = (y);(x) (z);--(x) ) const int MAXN = + +; int N,ARR[MAXN * 2]; Arr saves the index that has occurred int dfs (int cur,int curd,const int & maxd) { if (curd = = Maxd) return arr[cur-1] = = N; //The current value can be up to square (1<< (Maxd-curd)) times, after calculating so many times, is less than n if (arr[cur-1] * 1<< (Maxd-curd) < n) return 0; //Enumerate indices that have occurred FORD (i,cur-1,0) { //Generate new value, go to next layer Arr[cur] = arr[cur-1] + arr[i]; if (DFS (cur + 1,curd + 1, maxd)) return 1;Arr[cur] = arr[cur-1]-arr[i]; if (DFS (cur + 1,curd + 1, maxd)) return 1;        } return 0;}int init () { int tmp = n,cnt = 0;While (tmp > 1) {tmp/= 2;++cnt;        } return cnt;}int Main () { //freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout); while (~scanf ("%d", &n) && N) {For (int i = init ();; ++i) {arr[0] = 1; if(Dfs (1,0,i)) {printf ("%d\n", i); break;                         }                }                      } return 0;}

From for notes (Wiz)

[2016-03-03] [UVA] [1374] [Power calculus]