埃及分数问题(迭代加深)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL tmpans[100000],ans[100000];
 5 int maxd;
 6 /**
 7  *学习:1.memcpy:复制结构体时加&,
 8              数组时不加,数组名即地址
 9         2.迭代加深搜索,剪枝条
10  * */
11 int get_maxb(LL a,LL b){
12     //
13     //for(int i = 1 ; ; i ++){
14     //    if(a*i >= b) return i;
15     //}
16     return b/a+1;  
17 }
18 bool is_better(int d){
19     for(int i = d ; i >= 0 ; i --) if(tmpans[i] != ans[i]){
20         return ans[i] == -1 || tmpans[i] < ans[i];
21     }
22     return false;
23 }
24 
25 bool dfs(int d,int from,LL a,LL b){
26     if(d == maxd){
27         if(b%a) return false;
28         tmpans[d] = b / a;
29         if(is_better(d)) memcpy(ans,tmpans,sizeof(LL) * (d+1));
30         return true;
31     }
32     //
33     bool ok = false;
34     from = max(from,get_maxb(a,b));
35     for(int i = from ; ; i ++){
36         // 剪枝
37         if(a*i >= (maxd-d+1) * b) break; 
38         tmpans[d] = i;
39         // 剩下的分数
40         LL bb = b*i;
41         LL aa = a*i - b;
42         LL g = __gcd(aa,bb);
43         if(dfs(d+1,i+1,aa/g,bb/g)) ok = true;
44     }
45     return ok;
46 }
47 
48 int main(){
49     LL a,b;
50     int kase = 1;
51     while(scanf("%lld%lld",&a,&b) != EOF && (a || b)){
52         bool ok = false;
53         memset(ans,-1,sizeof(ans));
54         for(maxd = 1 ; ; maxd ++){
55             if(dfs(0,get_maxb(a,b),a,b)) { ok = true; break; }
56         }
57         printf("Case %d:%lld/%lld=",kase++,a,b);
58         for(int i = 0 ; i < maxd ; i ++) printf("%d/%lld+",1,ans[i]);
59         printf("%d/%lld\n",1,ans[maxd]);
60     }
61     return 0;
62 }