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,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:
x15 = x14 x x, x30 = x15 x x15, x31 = x30 x x.
x2 = x x x, x4 = x2 x x2, x8 = x4 x x4, x10 = x8 x x2,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.
x20 = x10 x x10, x30 = x20 x x10, x31 = x30 x x.
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
1 31 70 91 473 512 811 953 0
Sample Output
0 6 8 9 11 9 13 12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;
}