hdoj GTW likes math 5595 （暴力）

GTW likes math

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 70 Accepted Submission(s): 44

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

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

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

Hint

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

//有个坑点，是所有整数域的最值，所以只用考虑整数就行了。

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
	int t;
	int a,b,c,s,e;
	int mi,ma;
	int sum;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d%d%d",&a,&b,&c,&s,&e);
		mi=0x3f3f3f3f;ma=-0x3f3f3f3f;
		for(int i=s;i<=e;i++)
		{
			sum=a*i*i+b*i+c;
			if(sum>ma)
				ma=sum;
			if(sum<mi)
				mi=sum;
		}
		printf("%d %d\n",ma,mi);
	}	
	return 0;
}