埃及分数 迭代加深搜索 IDA*

迭代加深搜索 IDA*

首先枚举当前选择的分数个数上限maxd，进行迭代加深

之后进行估价，假设当前分数之和为a,目标分数为b，当前考虑分数为1/c，那么如果1/c×（maxd - d）< a - b那么不可能得解，进行剪枝

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 
 const int maxn = ;
 const int inf = ;
 int a, b;
 int ans[maxn], cur[maxn];
 int maxd = ;
 
 bool better(void) {
     for (int i = maxd; i >= ; i--) {
         if (cur[i] < ans[i]) return ();
         else if (cur[i] > ans[i]) return ();
     }
 }
 
 int gcd(int x, int y) {
     return (y ==  ? x : gcd(y, x % y));
 }
 
 int getfirst(int aa, int bb) {
     if (bb % aa == ) return (bb / aa);
     else return (bb / aa + );
 }
 
 void dfs(int d, int aa, int bb) {
     if (d == maxd + ) {
         if (!aa && better()) memcpy(ans, cur, sizeof(int) * (maxd + ));
         return;
     }
     int fr = std :: max(getfirst(aa, bb), cur[d-] + );
     for (; ; fr++) {
         if (aa * fr > (bb * (maxd - d + ))) break;
         cur[d] = fr;
         int g = gcd(bb, fr);
         int ta = (aa * fr - bb) / g;
         int tb = (bb * fr / g);
         dfs(d + , ta, tb);
     }
 }
 int main () {
     scanf("%d %d", &a, &b);
     for (maxd = ; ; maxd++) {
         memset(ans, , sizeof(ans));
         dfs(, a, b);
         if (ans[] >  && ans[] != inf) {
             int cur = ;
             while (ans[cur] >  && ans[cur] != inf) {
                 printf("%d ", ans[cur]);
                 cur++;
             }
             break;
         }
     }
 
     return ;
 }