uva 12558 Egyptian Fractions (HARD version)

迭代加深dfs 。數據要用long long巨坑

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<stack>
#include<sstream>
using namespace std;


int maxn, ok, a, b, k;
long long c[100000], ans[100000];
set<long long> ban;
long long gcb(long long a, long long b) {
	return b == 0 ? a : gcb(b, a%b);
}
bool better(int cur)
{
	for (int i = cur; i >= 0; i--) {
		if (ans[i] != -1 && c[i] != ans[i])
		{
			return  c[i] < ans[i];
		}
	}
	return true;
}
void dfs(int cur, long long start, long long a, long long b) {
	if (cur == maxn - 1) {
		int m = gcb(a, b);
		a /= m;
		b /= m;
		c[cur] = b;
		if (a == 1 && !ban.count(b))
		{
			ok = 1;
			if (better(cur))
				memcpy(ans, c, sizeof(long long)*(cur + 1));

		}
		return;
	}
	start = max(start, b / a + 1);
	for (long long i = start;; i++) {
		if (ban.count(i))	continue;
		if ((maxn - cur)*b <= a * i)
			return;
		long long c1 = a * i - b;
		long long d1 = b * i;
		int m = gcb(c1, d1);
		c1 /= m;
		d1 /= m;
		c[cur] = i;
		dfs(cur + 1, i + 1, c1, d1);
	}
}
int main() {
	
	int t, kase = 0;
	cin >> t;
	while (t--) {
		cin >> a >> b >> k;
		ban.clear();
		ok = 0;
		while (k--) {
			long long temp;
			cin >> temp;
			ban.insert(temp);
		}
		for (maxn = 2;; maxn++) {
			memset(ans, -1, sizeof(ans));
			dfs(0, b / a + 1, a, b);
			if (ok)
				break;
		}

		printf("Case %d: %d/%d=", ++kase, a, b);
		for (int i = 0; i < maxn; i++)
		{
			printf("1/%lld", ans[i]);
			if (i != maxn - 1)	printf("+");
			else
			{
				printf("\n");
			}
		}
	}
	return 0;
}

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<stack>
#include<sstream>
//#include <Eigen/Dense>
//#include <Eigen/Geometry>
//using namespace Eigen;
using namespace std;



typedef long long LL;
int a, b, k, maxd, ok;
const int maxn = 1e5 + 10;
LL v[maxn], ans[maxn];//v爲dfs時的中間
set<LL> ban;

LL gcd(LL a, LL b)   //輾轉相除法，求最大公因子
{
	return b == 0 ? a : gcd(b, a%b);
}
// 返回滿足1/c <= a/b的最小c
LL get_first(LL a, LL b)
{
	return b / a + 1;
}

// 如果當前解v比目前最優解ans更優，更新ans
bool better(int d)  //前d個，這時d只能是maxd，
{
	for (int i = d; i >= 0; i--) 
		if (v[i] != ans[i])
			return ans[i] == -1 || v[i] < ans[i];//沒被訪問過，或者是當前序列更優(因爲題目要求數量相同時，最小分數越大越好)
	return false;
}
// 當前深度爲d，分母(注意是分母)不能小於from，分數之和恰好爲aa/bb
void dfs(int cur, LL from, LL aa, LL bb)
{
	if (cur == maxd)
	{
		if (bb % aa || ban.count(bb/aa)) 
			return; // aa/bb必須是埃及分數（任何數mod1都等於0，可以起到一部分剪枝的作用）
		v[cur] = bb / aa;
		if (better(cur))
		{
			memcpy(ans, v, sizeof(LL)*(cur + 1));
			ok = 1;
		}
		return;
	}
	from = max(from, get_first(aa, bb));
	for (int i = from; ; i++)  //進行枚舉
	{
		// 剪枝：如果剩下的maxd+1-d個分數全部都是1/i，加起來仍然不超過aa/bb，則無解
		if (aa*i >= bb * (maxd - cur + 1)) return;//從第0層開始算的，總共有maxn+1個
		if (ban.count(i))
			continue;
		v[cur] = i;
		// 計算aa/bb - 1/i，設結果爲a2/b2
		LL a1 = aa * i - bb;
		LL b1 = bb * i;
		LL g = gcd(a1, b1); // 以便約分
        //除以最大公因數，防止下次計算aa*i和 bb * (maxd - cur + 1)溢出。此題數據實際不會溢出
		dfs(cur + 1, i + 1, a1 / g, b1 / g);
		//dfs(cur + 1, i + 1, a1, b1);
	}
	return;//注意這些返回條件
}
int main()
{
	int kase = 0;
	int cnt;
	cin >> cnt;
	while (cnt--)
	{
		ban.clear();
		cin >> a >> b >> k;
		while (k--)
		{
			LL temp;
			cin >> temp;
			ban.insert(temp);
		}
		ok = 0;
		for (maxd = 1; ; maxd++) //從小到大，枚舉深度上限
		{
			memset(ans, -1, sizeof(ans));
			dfs(0, get_first(a, b), a, b);
			if (ok)
				break;
		}
		cout << "Case " << ++kase << ": ";
		cout << a << "/" << b << "=";
		for (int i = 0; i < maxd; i++) 
			cout << "1/" << ans[i] << "+";
		cout << "1/" << ans[maxd] << "\n";	
	}
	return 0;
}