UVa 1374 Power Calculus (IDA*或都打表)
Posted dwtfukgv
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题意:给定一个数n,让你求从1至少要做多少次乘除才可以从 x 得到 xn。
析:首先这个是幂级的,次数不会很多,所以可以考虑IDA*算法,这个算法并不难,难在找乐观函数h(x),
这个题乐观函数可以是当前最大数*2maxd - d 小于n,回溯。很好理解,最大的数再一直乘2都达不到,最终肯定达不到。
再就是应该先试乘再试除,还有不要出现负整数。我测了不少知道应该是13次最多,所以这也是一个优化。
为了追求速度,也可以先1~1000的数打表。
代码如下:
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <set> #include <vector> #include <algorithm> using namespace std; const int maxn = 1000 + 5; int maxd, n, a[maxn]; bool dfs(int d, int num, int cur){ if(d == maxd){ if(n == cur) return true; return false; } int m = 0; for(int i = 0; i < num; ++i) m = max(m, a[i]); if(m * (1 << (maxd - d)) < n) return false;//回溯 for(int i = num-1; i >= 0; --i){ a[num] = cur + a[i]; ++num;//乘法 if(dfs(d+1, num, cur+a[i])) return true; a[num-1] = abs(cur - a[i]);//除法 if(dfs(d+1, num, abs(cur-a[i]))) return true; --num; } return false; } int main(){ while(scanf("%d", &n) == 1 && n){ for(maxd = 0; maxd < 13; ++maxd){ a[0] = 1; if(dfs(0, 1, 1)) break; } printf("%d\n", maxd); } return 0; }
下面是打表的代码:
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <set> #include <vector> #include <algorithm> using namespace std; const int ans[]={0,1,2,2,3,3,4,3,4,4,5,4,5,5,5,4,5,5,6,5,6,6,6,5,6,6,6,6,7,6,6,5,6,6, 7,6,7,7,7,6,7,7,7,7,7,7,7,6,7,7,7,7,8,7,8,7,8,8,8,7,8,7,7,6,7,7,8,7,8,8,8,7,8,8,8,8, 8,8,8,7,8,8,8,8,8,8,9,8,9,8,9,8,8,8,8,7,8,8,8,8,9,8,9,8,9,9,9,8,9,9,9,8,9,9,9,9,9,9,9,8, 9,9,9,8,9,8,8,7,8,8,9,8,9,9,9,8,9,9,9,9,9,9,9,8,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9, 8,9,9,9,9,9,9,10,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,9,10,9,10,9,9,9,9,8, 9,9,9,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,10,10,10, 10,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,9,10,10,10,10,10,10,10,9,10,10, 10,9,10,9,9,8,9,9,10,9,10,10,10,9,10,10,11,10,11,10,10,9,10,10,11,10,11,10,10,10, 10,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,10,10,11,10,11,11,11,10,11,10,11,10, 11,10,11,10,11,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,11,10,11,10,11,11,11,10,11,11,11, 10,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11, 10,11,11,11,10,11,11,11,10,11,10,11,10,10,10,10,9,10,10,10,10,11,10,11,10,11,11,11,10,11, 11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11, 11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11, 11,11,11,11,11,11,11,12,11,12,11,11,11,12,11,12,11,11,11,11,11,11,11,11,11,11,10,11,11, 11,11,11,11,11,11,11,11,12,11,12,11,11,10,11,11,12,11,12,11,11,10,11,11,11,10,11,10,10,9,10, 10,11,10,11,11,11,10,11,11,12,11,12,11,11,10,11,11,12,11,12,12,11,11,12,12,12,11,12,11,11,10, 11,11,12,11,12,12,12,11,11,12,12,11,12,11,12,11,11,11,12,11,12,11,11,11,12,11,11,11,11,11,11,10,11, 11,11,11,11,11,12,11,11,11,12,11,12,12,12,11,12,11,12,11,12,12,12,11,12,12,12,12,12, 12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,11,12,11, 12,11,11,11,11,11,11,10,11,11,11,11,11,11,12,11,12,11,12,11,12,12, 12,11,12,12,12,11,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12, 12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12, 12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12, 12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,12, 11,12,11,11,11,11,10,11,11,11,11,12,11,12,11,12,12,12,11,12,12,12,11,12,12,12,12, 12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,12,12, 12,11,12,12,12,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,13,12,12,12,12,12,12,11,12,12,12,12,12, 12,13,12,12,12,13,12,13,12,12,12,13,12,13,12,13,12,12,12,12,12,12,12,12,12, 12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12,13,12,13,12,13,12,13,12,13,12,13, 13,13,12,13,13,13,12,13,12,13,12,13,13,13,12,13,13,13,12,13,12,13,12,12,12,13,12, 13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12, 12,12,12,12,13,12,13,13,13,12,13,13,13,12,13,12,12,11,12,12,13,12,13,13,13,12}; int main() { int n; while(scanf("%d",&n) == 1 && n) printf("%d\n", ans[n-1]); return 0; }
可以看到最大才是13,次数不会很多。
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