hdu 5742 It's All In The Mind（2016多校第二场）

It's All In The Mind

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 505    Accepted Submission(s): 225

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

 

题意：求a1+a2/a1+a2+..+an的最大值，其中只给出部分的值，ax=y;

 

水题，保证a1,a2最大，后面的尽量小就好，需要注意的是，这是非递增数列，后面有数赋值，会影响前面的数的取值。

 

附上代码：

 

#include <iostream>
#include <cstdio>
using namespace std;
int a[105];
int _gcd(int x,int y) ///求最大公约数
{
    int z;
    if(x<y) z=x,x=y,y=z;
    while(y)
    {
        z=x%y;
        x=y;
        y=z;
    }
    return x;
}
int main()
{
    int T,i,j,n,m;
    scanf("%d",&T);
    while(T--)
    {
        int x,y,t=0;
        scanf("%d%d",&n,&m);
        for(i=1; i<=n; i++)  ///初始化全为-1
            a[i]=-1;
        for(i=0; i<m; i++)
        {
            scanf("%d%d",&x,&y);
            a[x]=y;
        }
        if(n==2||m==0)   ///若只有两个数，或者不给任何数赋值，都能输出最大的1/1
        {
            printf("1/1\n");
            continue;
        }
        int nn=0,mm=0,fail=0;
        for(i=n; i>=3; i--)  ///因为前面的小于等于后面的，因此从后面向前面循环
        {
                if(a[i]!=-1) ///如果a[i]被赋值，记录此时的a[i]，并标记已出现赋值的数
                {
                    t=a[i];
                    fail=1;
                    mm+=t;
                }
                else
                {
                    if(!fail)  ///若没有出现已赋值的数，可以定义为后面的数始终为0
                    {
                        mm+=0;
                    }
                    else   ///若已出现，则只能小于等于这个数
                    {
                        mm+=t;
                    }
                }
        }
        if(a[1]==-1) ///a1和a2期望越大越好，若都无赋值，就都取100，若有赋值，则去赋值的数，且a2小于a1
        {
            nn+=100,mm+=100;
            if(a[2]==-1)
                nn+=100,mm+=100;
            else
                nn+=a[2],mm+=a[2];
        }
        else
        {
            nn+=a[1],mm+=a[1];
            if(a[2]==-1)
                nn+=a[1],mm+=a[1];
            else
                nn+=a[2],mm+=a[2];
        }
        int dd=_gcd(nn,mm);
        printf("%d/%d\n",nn/dd,mm/dd);
    }
    return 0;
}