埃及分数

埃及分数，题目参考http://acm.fjnu.edu.cn/show?problem_id=1341

题目：

在古埃及，人们使用单位分数的和(形如1/a的, a是自然数)表示一切有理数。
如：2/3=1/2+1/6,但不允许2/3=1/3+1/3,因为加数中有相同的。
对于一个分数a/b,表示方法有很多种，但是哪种最好呢？
首先，加数少的比加数多的好，其次，加数个数相同的，最小的分数越大越好。
如：

19/45=1/3 + 1/12 + 1/180

19/45=1/3 + 1/15 + 1/45

19/45=1/3 + 1/18 + 1/30,

19/45=1/4 + 1/6 + 1/180

19/45=1/5 + 1/6 + 1/18.

最好的是最后一种，因为1/18比1/180,1/45,1/30,1/180都大。

给出a,b(0〈a〈b〈1000),编程计算最好的表达方式。

 

想法：

使用IDA*算法实现，按照分母从小到大的顺序搜索，外加剪枝。测试了几组数据，感觉是正确的。开始时提交显示TLE。于是我将动态创建的vecotr，改成在创建时就分配。然后再接再厉，结果提交了几次都是Wrong Answer，真是郁闷。先贴上来，改天再想！

 #include <iostream> #include <algorithm>#include <cmath>#include <vector>#include <cassert>using namespace std; const double EPSION = 1e-30;namespace Math {static int GCD(int a, int b){if (a < b){int tmp = a;a = b;b = tmp;}if (a % b == 0)return b;elsereturn GCD(b, a % b);}};struct Fraction {Fraction() {}Fraction(int n, int d) : numerator(n), denominator(d) { }int numerator;int denominator; //Fraction operator-(const Fraction &lhs){int nld = numerator * lhs.denominator;return Fraction(nld - lhs.denominator * denominator, nld);}inline void normalize();};///将分子与分母规约成最简式void Fraction::normalize(){int gcd = Math::GCD(denominator, numerator);numerator /= gcd;denominator /= gcd;}class Problem{public:Problem();void Solve();void IdaIterator(int curStep, int totalStep);void IdaStar();void Input(int numerator, int denominator);void Output();private:static const int MAXSTEP = 1000;Fraction destFraction;double destValue;int bestStep;bool success; //找到一个可行解std::vector <int> bestX; //最优解的分母std::vector <int> curX; //当前解的分母std::vector <Fraction> diff; //当前解对应的差值};Problem::Problem(){success = false;bestStep = MAXSTEP;bestX.resize(MAXSTEP + 1);diff.resize(MAXSTEP + 1);curX.resize(MAXSTEP + 1);}void Problem::Solve(){success = false;bestStep = MAXSTEP;destFraction.normalize();if (destFraction.numerator == 1){bestStep = 1;bestX[1] = destFraction.denominator;}else{IdaStar();}}void Problem::Input(int numerator, int denominator){destFraction.denominator = denominator;destFraction.numerator = numerator;}void Problem::Output(){for (int i = 1; i < bestStep; ++i){cout << bestX[i] << " ";}cout << bestX[bestStep] << endl;}void Problem::IdaStar(){int steps = 2;while (!success){curX[0] = destFraction.denominator / destFraction.numerator;diff[0] = destFraction;// - Fraction(1, curX[0]);IdaIterator(1, steps++);}}//curStep must be >= 2//第curStep个数为v//当前差值为d (diff[curStep-1].toDouble())//显然1/v <= diff[curStep-1] => v >= 1 / d, 另外根据题意v >= x[curStep-1] + 1//如果还有n步，那么n * 1/v >= d => v <= n / d//另外根据题意v <= *bestx.rbegin() - 1//因此，max(x[curStep-1] + 1, 1/d) <= v <= min(n / d, *bestx.regin() - 1)void Problem::IdaIterator(int curStep, int totalStep){if (curStep > totalStep){if (diff[totalStep].numerator == 0){success = true;if (totalStep < bestStep || (totalStep == bestStep && curX[totalStep] < bestX[totalStep])){bestStep = totalStep;for (int i = 1; i <= totalStep; ++i){bestX[i] = curX[i];}}}return;}int startPos = diff[curStep-1].denominator / diff[curStep-1].numerator; // 1/d//显然如果还少c/d,那么增加的值t必须 < c/d, c/d < 1 / [d/c]//所以t < 1 /[d/c]if (startPos < curX[curStep - 1] + 1)startPos = curX[curStep - 1] + 1;int lastPos = (totalStep - curStep + 1) * diff[curStep - 1].denominator / diff[curStep - 1].numerator;if (success){//要成为最优解，其最大的分母要小于最优解的最大分母if (lastPos > bestX[bestStep])lastPos = bestX[bestStep];}for (int v = startPos; v <= lastPos; ++v){curX[curStep] = v;if (diff[curStep - 1].numerator * v >= diff[curStep - 1].denominator){Fraction fr;fr.denominator = diff[curStep - 1].denominator * v;fr.numerator = diff[curStep - 1].numerator * v - diff[curStep - 1].denominator;diff[curStep] = fr;IdaIterator(curStep + 1, totalStep);}}}int main(int argc, char *argv[]) { int n, c, d;cin >> n;Problem pro;while (n-- > 0){scanf("%d %d", &c, &d);pro.Input(c, d);pro.Solve();pro.Output();}return 0; }

 

后记：

无意间测试了一组数据，发现进入死循环，经过排查，发现是数据精度的问题，如果只用int，那么就会溢出，从而变成负数。

所以使用了double类型，再次提交，总算成功了。

#include <iostream> #include <algorithm>#include <cmath>#include <vector>#include <ctime>#include <cassert>using namespace std; const double EPSION = 1e-30;namespace Math {static int GCD(int a, int b){if (a < b){int tmp = a;a = b;b = tmp;}if (a % b == 0)return b;elsereturn GCD(b, a % b);}};struct Fraction {Fraction() {}//Fraction(int n, int d) : numerator(n), denominator(d) { }Fraction(double n, double d) : numerator(n), denominator(d) { }double numerator;double denominator; //使用int会导致精度不够inline void normalize();};///将分子与分母规约成最简式void Fraction::normalize(){if (fabs(numerator) == 0 || fabs(denominator) == 0)return;int gcd = Math::GCD(static_cast<int>(denominator), static_cast<int>(numerator));numerator /= gcd;denominator /= gcd;}class Problem{public:Problem();void Solve();void IdaIterator(int curStep, int totalStep);void IdaStar();void Input(int numerator, int denominator);void Output();private:static const int MAXSTEP = 1000;Fraction destFraction;double destValue;int bestStep;bool success; //找到一个可行解std::vector <int> bestX; //最优解的分母std::vector <int> curX; //当前解的分母std::vector <Fraction> diff; //当前解对应的差值};Problem::Problem(){success = false;bestStep = MAXSTEP;bestX.resize(MAXSTEP + 1);diff.resize(MAXSTEP + 1);curX.resize(MAXSTEP + 1);}void Problem::Solve(){success = false;bestStep = MAXSTEP;destFraction.normalize();if (destFraction.numerator == 1){bestStep = 1;bestX[1] = destFraction.denominator;}else{IdaStar();}}void Problem::Input(int numerator, int denominator){destFraction.denominator = denominator;destFraction.numerator = numerator;}void Problem::Output(){for (int i = 1; i < bestStep; ++i){cout << bestX[i] << " ";}cout << bestX[bestStep] << endl;}void Problem::IdaStar(){int steps = 2;while (!success){curX[0] = destFraction.denominator / destFraction.numerator;diff[0] = destFraction;// - Fraction(1, curX[0]);IdaIterator(1, steps++);}}//curStep must be >= 2//第curStep个数为v//当前差值为d (diff[curStep-1].toDouble())//显然1/v <= diff[curStep-1] => v >= 1 / d, 另外根据题意v >= x[curStep-1] + 1//如果还有n步，那么n * 1/v >= d => v <= n / d//另外根据题意v <= *bestx.rbegin() - 1//因此，max(x[curStep-1] + 1, 1/d) <= v <= min(n / d, *bestx.regin() - 1)void Problem::IdaIterator(int curStep, int totalStep){if (curStep > totalStep){if (fabs(diff[curStep - 1].numerator) < EPSION){success = true;if (totalStep < bestStep || (totalStep == bestStep && curX[totalStep] < bestX[totalStep])){bestStep = totalStep;for (int i = 1; i <= totalStep; ++i){bestX[i] = curX[i];}}}return;}assert(diff[curStep-1].numerator != 0);int startPos = static_cast<int>(diff[curStep-1].denominator / diff[curStep-1].numerator); // 1/d//显然如果还少c/d,那么增加的值t必须 < c/d, c/d < 1 / [d/c]//所以t < 1 /[d/c]if (startPos < curX[curStep - 1] + 1)startPos = curX[curStep - 1] + 1;int lastPos = static_cast<int>((totalStep - curStep + 1) * diff[curStep - 1].denominator / diff[curStep - 1].numerator);if (success){//要成为最优解，其最大的分母要小于最优解的最大分母if (lastPos > bestX[bestStep])lastPos = bestX[bestStep];}for (int v = startPos; v <= lastPos; ++v){assert(v >= 0);curX[curStep] = v;if (diff[curStep - 1].numerator * v >= diff[curStep - 1].denominator){Fraction fr;fr.denominator = diff[curStep - 1].denominator * v;fr.numerator = diff[curStep - 1].numerator * v - diff[curStep - 1].denominator;//if (fr.denominator < 0)//{//cout << diff[curStep - 1].denominator << " " << v;//system("PAUSE");//}//assert (fr.numerator >= 0);//fr.normalize();//在此处不应该调用normalize，因为已经使用了浮点数，直接截断int数就会出现负数diff[curStep] = fr;IdaIterator(curStep + 1, totalStep);}}}int main(int argc, char *argv[]) { int n, c, d;cin >> n;Problem pro;while (n-- > 0){scanf("%d %d", &c, &d);//clock_t start = clock();pro.Input(c, d);pro.Solve();pro.Output();//clock_t end = clock();//cout << "time cost = " << end - start << endl;}return 0; }