fix(search/idastar): 埃及分数部分样例卡死 (#5364)

Co-authored-by: Tifa <62847935+Tiphereth-A@users.noreply.github.com>

docs/search/code/idastar/idastar_1.cpp

@@ -1,71 +1,66 @@
-// 埃及分数问题
-#include <algorithm>
-#include <cassert>
-#include <cstdio>
-#include <cstring>
-#include <iostream>
+#include <bits/stdc++.h>
+
 using namespace std;
 
-int a, b, maxd;
+const int MAX_E = 1e7;
+int a, b;
+vector<int> ans;
+vector<int> current;
 
-long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
+inline bool better() { return ans.empty() || current.back() < ans.back(); }
 
-// 返回满足 1/c <= a/b 的最小 c 值
-int get_first(long long a, long long b) { return b / a + 1; }
+bool dfs(int d, long a, long b, int e) {
+  if (d == 0) {
+    if (a == 0 && better()) ans = current;
+    return a == 0;
+  }
 
-const int maxn = 100 + 5;
+  long _gcd = gcd(a, b);
+  a /= _gcd;
+  b /= _gcd;
 
-long long v[maxn], ans[maxn];
-
-// 如果当前解 v 比目前最优解 ans 更优，更新 ans
-bool better(int d) {
-  for (int i = d; i >= 0; i--)
-    if (v[i] != ans[i]) {
-      return ans[i] == -1 || v[i] < ans[i];
+  bool solved = false;
+  // the min value of e:
+  // a/b - 1/e >= 0
+  // e >= b/a
+  int e1 = max(e + 1, int((b + a - 1) / a));
+  // b/a <= e <= MAX_E
+  // b/a <= MAX_E
+  if (b > a * MAX_E) {
+    return false;
+  }
+  for (;; e1++) {
+    // the max value of e:
+    // d * (1/e) >= a/b
+    // d/e >= a/b
+    if (d * b < a * e1) {
+      return solved;
     }
-  return false;
+    current.push_back(e1);
+    // a/b - 1/e
+    solved |= dfs(d - 1, a * e1 - b, b * e1, e1);
+    current.pop_back();
+  }
+  return solved;
 }
 
-// 当前深度为 d，分母不能小于 from，分数之和恰好为 aa/bb
-bool dfs(int d, int from, long long aa, long long bb) {
-  if (d == maxd) {
-    if (bb % aa) return false;  // aa/bb 必须是埃及分数
-    v[d] = bb / aa;
-    if (better(d)) memcpy(ans, v, sizeof(long long) * (d + 1));
-    return true;
-  }
-  bool ok = false;
-  from = max(from, get_first(aa, bb));  // 枚举的起点
-  for (int i = from;; i++) {
-    // 剪枝：如果剩下的 maxd+1-d 个分数全部都是 1/i，加起来仍然不超过
-    // aa/bb，则无解
-    if (bb * (maxd + 1 - d) <= i * aa) break;
-    v[d] = i;
-    // 计算 aa/bb - 1/i，设结果为 a2/b2
-    long long b2 = bb * i;
-    long long a2 = aa * i - bb;
-    long long g = gcd(a2, b2);  // 以便约分
-    if (dfs(d + 1, i + 1, a2 / g, b2 / g)) ok = true;
-  }
-  return ok;
+int solve() {
+  ans.clear();
+  current.clear();
+  for (int maxd = 1; maxd <= 100; maxd++)
+    if (dfs(maxd, a, b, 1)) return maxd;
+  return -1;
 }
 
 int main() {
   int kase = 0;
   while (cin >> a >> b) {
-    int ok = 0;
-    for (maxd = 1; maxd <= 100; maxd++) {  // 枚举深度上限
-      memset(ans, -1, sizeof(ans));
-      if (dfs(0, get_first(a, b), a, b)) {  // 开始进行搜索
-        ok = 1;
-        break;
-      }
-    }
+    int maxd = solve();
     cout << "Case " << ++kase << ": ";
-    if (ok) {
+    if (maxd > 0) {
       cout << a << "/" << b << "=";
-      for (int i = 0; i < maxd; i++) cout << "1/" << ans[i] << "+";
-      cout << "1/" << ans[maxd] << "\n";
+      for (int i = 0; i < maxd - 1; i++) cout << "1/" << ans[i] << "+";
+      cout << "1/" << ans[maxd - 1] << "\n";
     } else
       cout << "No solution.\n";
   }

docs/search/examples/idastar/idastar_1.ans

@@ -1 +1,2 @@
-Case 1: 495/499=1/2+1/5+1/6+1/8+1/3992+1/14970
+Case 1: 495/499=1/2+1/3+1/8+1/30+1/3992+1/14970
+Case 2: 3/997=1/345+1/14955+1/22931

docs/search/examples/idastar/idastar_1.in

@@ -1 +1,2 @@
-495 499
+495 499
+3 997