hdu 5742 It's All In The Mind

本文介绍了一种解决特定数学问题的算法,该问题要求在已知部分元素的情况下找到一个非递增序列,使得前两个元素之和与整个序列之和的比例最大化。通过使前两项尽可能大,剩余项和尽可能小来求解。

It's All In The Mind

#include<iostream>
#include<stdio.h>
using namespace std;
int gcd(int a , int b)
{

       if(b==0) return a;
          a%=b;
             return gcd(b,a);
}
int main(){
    int t;
    int num[1005];
    scanf("%d",&t);
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++) num[i]=-1;
        for(int i=0;i<m;i++ ){
            int tmp1,tmp2;
            scanf("%d%d",&tmp1,&tmp2);
            num[tmp1]=tmp2;
        }
        int sum=0;
        if(num[1]==-1) num[1]=100;
        if(num[2]==-1) num[2]=num[1];
        int tmp=0;
        for(int i=n;i>2;i--){
            if(num[i]==-1) num[i]=tmp;
            else{
                tmp=num[i];
            }
            sum+=num[i];
        }
        int div=gcd(sum+num[1]+num[2],num[1]+num[2]);
        if(div)
          printf("%d/%d\n",(num[1]+num[2])/div,(sum+num[1]+num[2])/div);
        else printf("0/1\n");
    }
    return 0;
}
View Code

首先判断 最后到结果是关于a1,a2的增函数,关于剩余和到减函数,令a1,a2尽可能大,其余数的和尽可能小就可以了。

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 565    Accepted Submission(s): 247


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}0ai100.
2. The sequence is non-increasing, i.e. a1a2...an.
3. The sum of all elements in the sequence is not zero.

Professor Zhang wants to know the maximum value of a1+a2ni=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 (2n100,0mn) -- the length of the sequence and the number of known elements.

In the next m lines, each contains two integers xi and yi (1xin,0yi100,xi<xi+1,yiyi+1), indicating that axi=yi.
 

 

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

 

Sample Input
2 2 0 3 1 3 1
 

 

Sample Output
1/1 200/201
 

 

Author
zimpha
 

 

Source
 

 

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转载于:https://www.cnblogs.com/superxuezhazha/p/5694631.html

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