HDU Problem 5615 Jam's math problem 【十字交叉，暴力】

Jam's math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1353    Accepted Submission(s): 610

Problem Description

Jam has a math problem. He just learned factorization.
He is trying to factorize ax2+bx+c into the form of pqx2+(qk+mp)x+km=(px+k)(qx+m).
He could only solve the problem in which p,q,m,k are positive numbers.
Please help him determine whether the expression could be factorized with p,q,m,k being postive.

 

Input

The first line is a number T, means there are T(1≤T≤100) cases 

Each case has one line,the line has 3 numbers a,b,c(1≤a,b,c≤100000000)

 

Output

You should output the "YES" or "NO".

 

Sample Input

2 1 6 5 1 6 4

 

Sample Output

YES NO

Hint

The first case turn $x^2+6*x+5$ into $(x+1)(x+5)$

 

Source

BestCoder Round #70

 

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#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
#define space " "
using namespace std;
//typedef long long LL;
typedef __int64 Int;
typedef pair<int,int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1);
const int MOD = 1e9+5;
const int MAXN = 1e5 + 10;
Int a, b, c;
int main() {
    int T;
    cin >> T;
    while (T--) {
        cin >> a >> b >> c;
        bool flag = false;
        if (b*b < 4*a*c) {
            cout << "NO" << endl;
        }
        else {
            Int sa = sqrt(a), sc = sqrt(c);
            for (int i = 1; i <= sa; i++) {
                if (a%i) continue;
                for (int j = 1; j <= sc; j++) {
                    if (c%j == 0) {
                        Int x = a/i, y = c/j;
                        if ((i*j + x*y) == b || (i*y + j*x) == b) {
                            flag = true; break;
                        }
                    }
                }
                if (flag) break;
            }
            if (flag)   cout << "YES" <<endl;
            else    cout << "NO" << endl;
        }
    }
    return 0;
}