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。