Sicily 2409. Egyptian Fractions

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很长的一道题目，说的是一个贪心减法的问题：对于一个M/N的分数，我们希望能够把它写成多个1/D相加的形式，并且每个D都要尽可能的大，不能超过1000000。题目已经说明要使用long long了，另外的话就按照贪心算法的一般思路去进行即可，无陷阱。

Run Time: 0.29sec

Run Memory: 304KB

Code Length: 1358Bytes

SubmitTime: 2012-03-02 12:58:39

// Problem#: 2409// Submission#: 1229318// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <cstdio>#include <vector>#include <stack>using namespace std;struct Fraction {    long long a, b;    void set( long long x, long long y ) { a = x; b = y; }    void subtract( const Fraction& f ) {        a = a * f.b - b * f.a;        b = b * f.b;        long long m = a, n = b, temp;        while ( m != 0 ) {            temp = m;            m = n % m;            n = temp;        }        a /= n;        b /= n;    }};int main(){    long long M, N;    Fraction target, sub;    while ( scanf( "%lld%lld", &M, &N ) && M && N ) {        stack<Fraction> s;        vector<Fraction> v;        target.set( M, N );        while ( !( target.a == 1 && target.b <= 1000000 ) ) {            if ( target.b <= 1000000 ) {                s.push( target );                sub.set( 1, target.b / target.a + 1 );                v.push_back( sub );                target.subtract( sub );            }            else {                target = s.top();                sub.set( 1, v.back().b + 1 );                v.pop_back();                v.push_back( sub );                target.subtract( sub );            }        }        v.push_back( target );        printf( "%lld", v[ 0 ].b );        for ( int i = 1; i < v.size(); i++ )            printf( " %lld", v[ i ].b );        printf( "\n" );    }    return 0;}