POJ3134 Power Calculus
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1. 题目描述
给定一个正整数$n$，求经过多少次乘法或除法运算可以从$x$得到$x^n$？中间结果也是可以复用的。
2. 基本思路
实际结果其实非常小，肯定不会超过20。因此，可以采用IDA*算法。
注意几个剪枝优化就好了：
（1）每次新计算的值必须从未出现过;
（2）每次新计算的值进行还可以执行的运算次数的幂运算仍然小于$x^n$，即新值左移还可以执行的次数小于$n$则一定不成立；
（3）该值与$n$的绝对值$\Delta$小于$n$，同时还可以执行的次数大于$ans[\Delta]+1$，那么一定成立。
3. 代码
（1）生成ans数组程序
1 /* 3134 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 #define INF 0x3f3f3f3f 44 #define mset(a, val) memset(a, (val), sizeof(a)) 45 46 const int maxn = 2048; 47 int ans[maxn]; 48 int a[maxn]; 49 bool visit[maxn]; 50 51 void printA(int *a) { 52 int sz = 1001; 53 54 rep(i, 0, sz) { 55 if (i == 0) { 56 printf("int ans[] = {%d", a[i]); 57 } else { 58 printf(", %d", a[i]); 59 } 60 } 61 printf("};\n"); 62 } 63 64 bool dfs(int dep, int step, int n) { 65 if (visit[n]) 66 return true; 67 68 if (dep > step) 69 return false; 70 71 if (a[dep-1]<<(step-dep+1) < n) 72 return false; 73 74 if (abs(n-a[dep-1])<n && step-dep>=ans[abs(n-a[dep-1])]) 75 return true; 76 77 int tmp; 78 79 rep(i, 0, dep) { 80 tmp = a[dep-1] + a[i]; 81 if (tmp<maxn && !visit[tmp]) { 82 visit[tmp] = true; 83 a[dep] = tmp; 84 if (dfs(dep+1, step, n)) 85 return true; 86 visit[tmp] = false; 87 } 88 tmp = a[dep-1] - a[i]; 89 if (tmp>0 && !visit[tmp]) { 90 visit[tmp] = true; 91 a[dep] = tmp; 92 if (dfs(dep+1, step, n)) 93 return true; 94 visit[tmp] = false; 95 } 96 } 97 98 return false; 99 } 100 101 int solve(int n) { 102 memset(visit, false, sizeof(visit)); 103 visit[1] = true; 104 visit[0] = true; 105 a[0] = 1; 106 rep(i, 1, n+1) { 107 if (dfs(1, i, n)) { 108 return i; 109 } 110 } 111 112 return -1; 113 } 114 115 void init() { 116 memset(ans, 0x3f, sizeof(ans)); 117 ans[1] = 0; 118 rep(i, 2, 1001) 119 ans[i] = solve(i); 120 printA(ans); 121 } 122 123 int main() { 124 ios::sync_with_stdio(false); 125 #ifndef ONLINE_JUDGE 126 freopen("data.in", "r", stdin); 127 freopen("data.out", "w", stdout); 128 #endif 129 130 init(); 131 132 #ifndef ONLINE_JUDGE 133 printf("time = %d.\n", (int)clock()); 134 #endif 135 136 return 0; 137 }
（2）打表程序
1 /* 3134 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 #define INF 0x3f3f3f3f 44 #define mset(a, val) memset(a, (val), sizeof(a)) 45 46 int ans[] = {1061109567, 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}; 47 48 int main() { 49 ios::sync_with_stdio(false); 50 #ifndef ONLINE_JUDGE 51 freopen("data.in", "r", stdin); 52 freopen("data.out", "w", stdout); 53 #endif 54 55 int n; 56 57 while (cin>>n && n) { 58 cout << ans[n] << endl; 59 } 60 61 #ifndef ONLINE_JUDGE 62 printf(
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