HDU1712周期

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5963    Accepted Submission(s):
3250

Problem Description

ACboy has N courses this term, and he plans to spend at
most M days on study.Of course,the profit he will gain from different course
depending on the days he spend on it.How to arrange the M days for the N courses
to maximize the profit?

 

Input

The input consists of multiple data sets. A data set
starts with a line containing two positive integers N and M, N is the number of
courses, M is the days ACboy has.
Next follow a matrix A[i][j],
(1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j
days on ith course he will get profit of value A[i][j].
N = 0 and M = 0 ends
the input.

 

Output

For each data set, your program should output a line
which contains the number of the max profit ACboy will gain.

 

Sample Input

2 2
1 2
1 3
2 2
2 1
2 1
2 3
3 2 1
3 2 1
0 0

 

Sample Output

3
4
6

 

Source

HDU
2007-Spring Programming Contest

 

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对于两个最简的分数 a / b, c / d 把他们两个的最小公倍数 x / y 也设为一个分数形式，那么这个 x 一定能够被 a , c整除， y 一定能够整除 b , d。那么要求得最小公倍数，那么肯定是分子尽量小，即 a , c 的最小公倍数， 分母尽量大， 即 b , d 的最大公约数。

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    if (b==) return a;
    else return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    ll i,j,t,fz1,fz2,fm1,fm2,fz,fm,len1,len2,len3,len4,tmp;
    char c;
    cin>>t;
    while (t--)
    {
        cin>>fz1>>c>>fm1;
        cin>>fz2>>c>>fm2;
        tmp=gcd(fz1,fm1);
        fz1/=tmp;
        fm1/=tmp;
        tmp=gcd(fz2,fm2);
        fz2/=tmp;
        fm2/=tmp;
        //cout<<tmp<<" "<<fz2<<" "<<fm2<<endl;
        //fm=gcd(fm1,fm2); 题目给的分数不是最简！！！！
        //fz=lcm(fz1,fz2);fz是俩分子的最小公倍数 后化简的话 fm已经变成两个数的最大公约数了 没法被用作化简了
        //cout<<gcd(fz,fm)<<"aaaaaaaaa"<<endl;
        //fz/=gcd(fz,fm);
        //fm/=gcd(fz,fm);//用abcd fz1,fz2写成fz1,fm1了
        if (gcd(fm1,fm2)==)
        cout<<lcm(fz1,fz2)<<endl;
        else
        cout<<lcm(fz1,fz2)<<"/"<<gcd(fm1,fm2)<<endl;
    }

    return ;
}

//很久之前做的很烦的一道题。