hdu 5742 It's All In The Mind（2016 Multi-University Training Contest 2——思维题）

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最新推荐文章于 2020-08-30 19:51:53 发布

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分类专栏： 2016 Multi-University Training Contest I believe i can

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2016 Multi-University Training Contest

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针对一个部分已知的非增序列，求特定分数表达式的最大值。通过优化未知元素值来实现目标函数最大化。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5742

It's All In The Mind

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 754 Accepted Submission(s): 335

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

(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

(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

Sample Output

Author

zimpha

Source

2016 Multi-University Training Contest 2

题目大意：求题目中所给这个公式的最大值，其中我们有一些数字我们不知道，他只会给部分数字，ax=y。（这个数列是递减的，可以等于）

解题思路：

对于分数，肯定是分母越大越好，分子越小越好，只要满足这种情况就会大。由于ai的范围是大于等于0，小于等于100的。所以ai的最大值就是100。

但是这个公式是递减的，所以后面的数字不能比前面给出的数字大，所以直接和前面相等即为最优解。

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

int a[110],Min;

int main()
{
    int t,sum;
    scanf("%d",&t);
    while (t--)
    {
        int n,m,k,x,y;
        k=sum=0;
        scanf("%d%d",&n,&m);
        memset(a,0,sizeof(a));
        for (int i=0; i<=2; i++)
            a[i]=100;
        for (int i=0; i<m; i++)
        {
            scanf("%d%d",&x,&y);
            a[x]=y;
        }
        if (a[2]>a[1])
            a[2]=a[1];
        for (int i=n;i>=3;i--)
        {
            if (a[i]!=0)
            {
                k=a[i];
                sum+=a[i];
                continue;
            }
            a[i]=k;
            sum+=a[i];
        }
        int mu=a[1]+a[2];
        int zi=sum+a[1]+a[2];
        int gcd=__gcd(mu,zi);
        printf ("%d/%d\n",mu/gcd,zi/gcd);
    }
    return 0;
}