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

31 4 40 0 11 3 1

Example Output

YESYESNO

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;}