UVA 1374 - Power Calculus(迭代深搜)

Starting with x and repeatedly multiplying by x, we can compute x³¹ with thirty multiplications:

x² = x x x,     x³ = x² x x,     x⁴ = x³ x x,     ...  ,     x³¹ = x³⁰ x x.

The operation of squaring can appreciably shorten the sequence of multiplications. The following is a way to compute x³¹ with eight multiplications:

x² = x x x,     x³ = x² x x,     x⁶ = x³ x x³,     x⁷ = x⁶ x x,     x¹⁴ = x⁷ x x⁷, 
x¹⁵ = x¹⁴ x x,     x³⁰ = x¹⁵ x x¹⁵,     x³¹ = x³⁰ x x.

This is not the shortest sequence of multiplications to compute x³¹. There are many ways with only seven multiplications. The following is one of them:

x² = x x x,     x⁴ = x² x x²,     x⁸ = x⁴ x x⁴,     x¹⁰ = x⁸ x x², 
x²⁰ = x¹⁰ x x¹⁰,     x³⁰ = x²⁰ x x¹⁰,     x³¹ = x³⁰ x x.

There however is no way to compute x³¹ with fewer multiplications. Thus this is one of the most efficient ways to compute x³¹ only by multiplications.

If division is also available, we can find a shorter sequence of operations. It is possible to compute x³¹with six operations (five multiplications and one division):

x² = x x x,     x⁴ = x² x x²,     x⁸ = x⁴ x x⁴,     x¹⁶ = x⁸ x x⁸,     x³² = x¹⁶ x x¹⁶,     x³¹ = x³²÷ x.

This is one of the most efficient ways to compute x³¹ if a division is as fast as a multiplication.

Your mission is to write a program to find the least number of operations to compute xⁿ by multiplication and division starting with x for the given positive integer n. Products and quotients appearing in the sequence of operations should be x to a positive integer's power. In other words, x^-3, for example, should never appear.

Input 

The input is a sequence of one or more lines each containing a single integer n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.

Output 

Your program should print the least total number of multiplications and divisions required to compute xⁿstarting with x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.

Sample Input 

13170914735128119530

Sample Output 

06891191312

题意：给定n，求出从1变换到n的最小步数，根据题意。

思路：迭代深搜。要剪枝。如果当前最大的不断自加到不了n，这种情况就没必要搜下去

代码：

#include <stdio.h>#include <string.h>#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)const int N = 55;const int mi[11] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024};int n, D, h[N], hn, ans[1001];void init() {for (int i = 0; i <= 10; i ++)if (n <= mi[i]) {D = i;break;}}bool dfs(int d, int sum) {if (d == D) {if (sum == n)return true;return false;}for (int i = hn - 1; i >= 0; i --) {int Max = 0;for (int j = 0; j < hn; j ++)Max = max(Max, h[j]);if (((Max + sum)<<(D - d - 1)) < n) return false;h[hn++] = sum + h[i];if (dfs(d + 1, sum + h[i])) return true;if (sum - h[i] > 0) {h[hn - 1] = sum - h[i];if (dfs(d + 1, sum - h[i])) return true;}hn--;}return false;}int solve() {for (;;D++) {memset(h, 0, sizeof(h));h[0] = 1; hn = 1;if (dfs(0, 1))break;}return D;}int main() {while (~scanf("%d", &n) && n) {init();printf("%d\n", solve());}return 0;}