hdu 5742

It's All In The Mind

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2300    Accepted Submission(s): 990

Problem Description

Professor Zhang has a number sequence a1,a2,...,an. However, the sequence is not complete and some elements are missing. Fortunately, Professor Zhang remembers some properties of the sequence:

1. For every i∈{1,2,...,n}, 0≤ai≤100.
2. The sequence is non-increasing, i.e. a1≥a2≥...≥an.
3. The sum of all elements in the sequence is not zero.

Professor Zhang wants to know the maximum value of a1+a2∑ni=1ai among all the possible sequences.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (2≤n≤100,0≤m≤n) -- the length of the sequence and the number of known elements.

In the next m lines, each contains two integers xi and yi (1≤xi≤n,0≤yi≤100,xi<xi+1,yi≥yi+1), indicating that axi=yi.

 

Output

For each test case, output the answer as an irreducible fraction " p/ q", where p, q are integers, q>0.

 

Sample Input

2 2 0 3 1 3 1

 

Sample Output

1/1 200/201

 

Author

zimpha

 

Source

2016 Multi-University Training Contest 2

 

思路：

        很明显，我们只需要分母尽量小，分子尽量大就可以了，很简单的贪心。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<queue>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
#define Max 3000100
#define MOD 530600414
typedef long long LL;
int t,n,m,arr[110];
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);
}
int main()
{
    cin>>t;
    while(t--)
    {
        int mark=0;
        memset(arr,-1,sizeof(arr));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            arr[a]=b;
        }
        for(int i=n;i>=3;i--)
        {
            if(arr[i]==-1)
                arr[i]=mark;
            else
                mark=arr[i];
        }
        if(arr[1]==-1)
            arr[1]=100;
        if(arr[2]==-1)
            arr[2]=arr[1];
        int fm=0,fz=0;
        for(int i=1;i<=n;i++)
            fm+=arr[i];
        fz=arr[1]+arr[2];
        int temp=gcd(fz,fm);
        printf("%d/%d\n",fz/temp,fm/temp);
    }
    return 0;
}