Equations(数学,水)

本文探讨了如何通过输入三个整数a、b、c来确定二次方程ax^2 + bx + c = 0的实数解数量,并提供了一个C++代码示例。包括了对a为0的特殊情况进行处理。

Equations

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 689    Accepted Submission(s): 315


Problem Description
All the problems in this contest totally bored you. And every time you get bored you like playing with quadratic equations of the form a*X 2 + b*X + c = 0. This time you are very curious to know how many real solutions an equation of this type has.
 

Input
The first line of input contains an integer number Q, representing the number of equations to follow. Each of the next Q lines contains 3 integer numbers, separated by blanks, a, b and c, defining an equation. The numbers are from the interval [-1000,1000].
 

Output
For each of the Q equations, in the order given in the input, print one line containing the number of real solutions of that equation. Print “INF” (without quotes) if the equation has an infinite number of real solutions.
 

Sample Input
  
3 1 0 0 1 0 -1 0 0 0
 

Sample Output
  
1 2 INF
 

Author
Mugurel Ionut Andreica
 

Source
 

Recommend
lcy
 

只要注意a为0的情况就ok

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define LL long long
#define maxi 2147483647
#define maxl 9223372036854775807
#define dg cout << "Here!" << endl;
using namespace std;

int main()
{
    int q, a, b, c, flag;
    cin >> q;
    while(q--)
    {
        cin >> a >> b >> c;
        if(!a)
        {
            if(!b)
            {
                if(!c) flag = 3;
                else flag = 0;
            }
            else flag = 1;
        }
        else
        {
            a = b * b - 4 * a * c;
            if(!a) flag = 1;
            else if(a > 0) flag = 2;
            else flag = 0;
        }
        if(!flag) puts("0");
        else if(flag == 1) puts("1");
        else if(flag == 2) puts("2");
        else puts("INF");
    }
    return 0;
}


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