GTW likes math
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 942 Accepted Submission(s): 426
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].
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.
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，给你一个方程a*x*x+b*x+c和一个取值范围【l，r】问当x在此区间内取值时，函数的最大值和最小值分别是多少
直接暴力求解
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 110
#define mod 10007
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
int t,Min,Max,i,j;
int a,b,c,l,r,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d%d%d",&a,&b,&c,&l,&r);
ans=0;Max=-INF;
Min=INF;
if(l>=0) //区间内全是正数
{
for(i=l;i<=r;i++)
{
ans=a*i*i+b*i+c;
Max=max(Max,ans);
Min=min(Min,ans);
}
}
else
{
//printf("%d*
",abs(l));
if(r>=0) //区间内有负有正
{
for(i=1;i<=abs(l);i++)
{
j=0-i;
//printf("%d*
",j);
ans=a*j*j+b*j+c;
Max=max(Max,ans);
Min=min(Min,ans);
}
for(i=0;i<=r;i++)
{
ans=a*i*i+b*i+c;
Max=max(Max,ans);
Min=min(Min,ans);
}
}
else//区间内全是负数
{
for(i=abs(r);i<=abs(l);i++)
{
j=0-i;
//printf("%d*
",j);
ans=a*j*j+b*j+c;
Max=max(Max,ans);
Min=min(Min,ans);
}
}
}
printf("%d %d
",Max,Min);
}
return 0;
}