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题目点此跳转

Starting with x and repeatedly multiplying by x, we can compute x 31 with thirty multiplications:

$x 2 = x \times x, x 3 = x 2 \times x, x 4 = x 3 \times x, . . ., x 31 = x 30 \times x .$
The operation of squaring can appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:
$x 2 = x \times x, x 3 = x 2 \times x, x 6 = x 3 \times x 3, x 7 = x 6 \times x, x 14 = x 7 \times x 7, x 15 = x 14 \times x, x 30 = x 15 \times x 15, x 31 = x 30 \times x .$
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:
$x 2 = x \times x, x 4 = x 2 \times x 2, x 8 = x 4 \times x 4, x 1 0 = x 8 \times x 2, x 20 = x 10 \times x 10, x 30 = x 20 \times x 10, x 31 = x 30 \times x .$
There however is no way to compute x31with 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 x31 with six operations (five multiplications and one division):
$x 2 = x \times x, x 4 = x 2 \times x 2, x 8 = x 4 \times x 4, x 16 = x 8 \times x 8, x 32 = x 16 \times x 16, x 31 = x 32 \div x .$
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 x n 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 x n starting 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

思路

题目意思是输入正整数n(1≤n≤1000),问最少需要几次乘除法可以从x得到xn ?例如,x31 需要6次:。计算过程中x的指数应当总是正整数(如x−3 =x/x4 是不允许的)。在

还是用IDA*, 将当前得到的指数集合作为状态, 转移就是指数的相加减, 但是不用枚举所有的指数相加减, 一种做法是每次用刚刚获得的那个数进行枚举其他的指数.
剪枝:d表示当前深度,maxd表示深度上限,则如果当前序列最大的数乘以2maxd−d 之后仍小于n,则剪枝.

代码

#include <algorithm>#include <iostream>#include <sstream>#include <utility>#include <string>#include <vector>#include <queue>#include <map>#include <set>#include <cstring>#include <cstdio>#include <cmath>#define met(a,b) memset(a, b, sizeof(a));#define IN freopen("in.txt", "r", stdin);#define OT freopen("out.txt", "w", stdout);using namespace std;typedef long long LL;typedef pair<int, int> PII;const int maxn = 1e4 + 10;const int INF = 0x7fffffff;const int dir[5][2] = {0,0,-1,0,1,0,0,-1,0,1};int n, maxd, a[maxn], ans[maxn];bool vis[maxn];bool dfs(int n, int deep) {    if(deep == maxd) {        if(a[deep] == n) return 1;        return 0;    }    for(int i = deep; i >= 0; --i) {        int t = a[i] + a[deep];        if(t * (1<<(maxd-deep)) < n) continue;        if(!vis[t]) {            vis[t] = 1; a[deep+1] = t;            if(dfs(n, deep+1)) return 1;            vis[t] = 0;        }        t = a[deep] - a[i];        if(t * (1<<(maxd-deep)) < n) continue;        if(!vis[t]) {            vis[t] = 1; a[deep+1] = t;            if(dfs(n, deep+1)) return 1;            vis[t] = 0;        }    }    return 0;}int IDAS(int n) {    if(n == 1) return 0;    for(maxd = (int)log2(n)-1; ; ++maxd) {        a[0] = 1; met(vis, 0); vis[1] = 1;        if(dfs(n, 0)) break;    }    return maxd;}int main() {    #ifdef _LOCAL    IN; //OT;    #endif // _LOCAL    while(scanf("%d", &n) == 1 && n) { printf("%d\n", IDAS(n)); }    return 0;}

另外这题如果不会剪枝的话,也可直直接跑出来结果, 不过要喝杯咖啡等一会,再将跑出来的数组直接交表
总共就1000个数.

#include <algorithm>#include <iostream>#include <sstream>#include <utility>#include <string>#include <vector>#include <queue>#include <map>#include <set>#include <cstring>#include <cstdio>#include <cmath>#define met(a,b) memset(a, b, sizeof(a));#define IN freopen("in.txt", "r", stdin);#define OT freopen("out.txt", "w", stdout);using namespace std;typedef long long LL;typedef pair<int, int> PII;const int maxn = 1e4 + 10;const int INF = 0x7fffffff;const int dir[5][2] = {0,0,-1,0,1,0,0,-1,0,1};int ans[] = {0,0, 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() {    #ifdef _LOCAL    IN; //OT;    #endif // _LOCAL    int n;    while(scanf("%d", &n) == 1 && n) { printf("%d\n", ans[n]); }    return 0;}