ACM 第八届山东省赛 F quadratic equation SDUT 3898

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quadratic equation

Time Limit: 2000MS Memory Limit: 131072KB

Submit Statistic

Problem Description

With given integers a,b,c, you are asked to judge whether the following statement is true: "For any x, if a⋅+b⋅x+c=0, then x is an integer."

Input

The first line contains only one integer T(1≤T≤2000), which indicates the number of test cases.
For each test case, there is only one line containing three integers a,b,c(−5≤a,b,c≤5).

Output

or each test case, output “YES” if the statement is true, or “NO” if not.

Example Input

3
1 4 4
0 0 1
1 3 1

Example Output

YES
YES
NO

Hint

Author

“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）

考察的是离散数学的蕴含式的理解   P-->Q     只有P为1 Q为0是 结果才是0  其余情况全为1  所以

当P为0 即P 不成立时 结果依然是YES  这是一个坑 

#include <iostream>
#include <algorithm>
#include <cmath>
#include <stdio.h>
#include <cstring>

using namespace std;

int main()
{
    int T;
    cin>>T;
    int a,b,c;
    int cont=1;
    while(T--)
    {
        cin>>a>>b>>c;
        int flag;
        if(a==0)
        {
            if(b!=0)
            {
                if(c==0)
                    flag=1;
                else if(c%b==0)
                    flag=1;
                else
                    flag=0;
            }
            else
            {
                if(c==0)
                    flag=0;
                else
                    flag=1;
            }
        }
        else
        {
            int detal=b*b-4*a*c;
            if(detal<0)
                flag=1;
            else
            {
                //printf("%d %lf\n",(int)sqrt(detal),sqrt(pow(sqrt(detal),2)));
                if((int)sqrt(detal)==sqrt(pow(sqrt(detal),2)))
                {
                    if((-b+(int)sqrt(detal))%(2*a)==0&&(-b-(int)sqrt(detal))%(2*a)==0)
                        flag=1;
                    else
                        flag=0;
                }
                else
                    flag=0;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}