HDOJ 5595 GTW likes math (暴力)

最新推荐文章于 2020-05-27 16:37:27 发布

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本文介绍了一道数学问题，即给定二次函数f(x)=ax²+bx+c，在整数区间[l,r]内找到该函数的最大值和最小值。通过遍历区间内的每个整数点并计算函数值来寻找最值。

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GTW likes math

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 316 Accepted Submission(s): 161

Problem Description

After attending the class given by Jin Longyu, who is a specially-graded teacher of Mathematics, GTW started to solve problems in a book titled “From Independent Recruitment to Olympiad”. Nevertheless, there are too many problems in the book yet GTW had a sheer number of things to do, such as dawdling away his time with his young girl. Thus, he asked you to solve these problems.

In each problem, you will be given a function whose form is like

f(x)=ax2+bx+c. Your assignment is to find the maximum value and the minimum value in the integer domain

[l,r].

Input

The first line of the input file is an integer

T, indicating the number of test cases. (

T≤1000)

In the following

T lines, each line indicates a test case, containing 5 integers,

a,b,c,l,r. (

|a|,|b|,|c|≤100,|l|≤|r|≤100), whose meanings are given above.

Output

In each line of the output file, there should be exactly two integers,

max and

min, indicating the maximum value and the minimum value of the given function in the integer domain

[l,r], respectively, of the test case respectively.

Sample Input

1
1 1 1 1 3

Sample Output

13 3
Hint
$f_1=3,f_2=7,f_3=13,max=13,min=3$

题意：求函数在区间内的最值
思路：数据不大，暴力就好

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define INF 0x7fffffff
#define LL __int64
using namespace std;
int a,b,c;
LL fun(double x)
{
    return a*x*x+b*x+c;
}
int main()
{
    int t,i;
    int low,high;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d",&a,&b,&c,&low,&high);
        LL ans1=-INF,ans2=INF;
        for(i=low;i<=high;i++)
        {
            LL k=fun(i);
            ans1=max(ans1,k);
            ans2=min(ans2,k);
        }
        printf("%I64d %I64d\n",ans1,ans2);
    }
    return 0;
}