BestCoder Round #66 1001

GTW likes math 

Accepts: 472      Submissions: 2140    

 Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others)

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)=ax^2​​+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 TT 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, maxmax and minmin, 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

f1=3,f2=7,f3=13,max=13,min=3

问题描述

某一天，GTW听了数学特级教师金龙鱼的课之后，开始做数学《从自主招生到竞赛》。然而书里的题目太多了，GTW还有很多事情要忙（比如把妹），于是他把那些题目交给了你。每一道题目会给你一个函数f(x)=ax^2+bx+cf(x)=ax​2​​+bx+c，求这个函数在整数区间[l,r]之间的最值。

输入描述

第一行一个整数T，表示数据组数。(T≤1000)
对于每一组数据，有一行，共五个整数a,b,c,l,r。(∣a∣≤100,∣b∣≤100,∣c∣≤100,∣l∣≤100,∣r∣≤100,l≤r)

输出描述

对于每一组数据，共一行两个整数max，min，表示函数在整数区间[l,r]中的最大值和最小值。

输入样例

1
1 1 1 1 2

输出样例

7 3

Hint

f​1​​=3,f​2​​=7,最大值=7,最小值=3

没必要用什么公式了，一路从L计算到R 保存最大最小就好

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int a,b,c;
int l,r;
int i,j;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>c>>l>>r;
        int maxn=-inf,minn=inf;
        for(i=l;i<=r;i++)
        {
            maxn=max(maxn,a*i*i+b*i+c);
            minn=min(minn,a*i*i+b*i+c);
        }
        cout<<maxn<<" "<<minn<<endl;
    }
    return 0;
}