HDU5615：Jam's math problem

最新推荐文章于 2018-08-22 12:13:30 发布

efine_dxq

最新推荐文章于 2018-08-22 12:13:30 发布

阅读量656

点赞数

CC 4.0 BY-SA版权

文章标签： acm HDU BastCoder math

本文链接：https://blog.youkuaiyun.com/efine_dxq/article/details/50615018

BestCoder 专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种解决特定形式的二次多项式因式分解问题的方法。提供了两种算法实现思路，一种是通过分解系数a和c的因子进行遍历匹配，另一种是利用判别式法判断是否存在正整数解。

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

方法一：

分解a、c因子，然后遍历。数据比较小，不会超时。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<cmath>
#include<vector>
#define ll long long
#define MAXN 1000000
struct node{
    int m, k;
}km[MAXN];
struct node1{
    int p, q;
}pq[MAXN];
int main(){
    int t;
    int a, b, c, cnta, cntb, flag;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d%d",&a, &b, &c);
        cnta= 0; cntb = flag = 0;
        for(int i = 1; i <= (int)sqrt(a); i ++){
            if(a%i==0) {
                pq[cnta].p = i;
                pq[cnta++].q = a/i;
            }
        }
        for(int i = 1; i <= (int)sqrt(c); i ++){
            if(c%i==0){
                km[cntb].k = i;
                km[cntb++].m = c/i;
            }
        }
        for(int i = 0; i < cnta; i++){
            int p, q;
            p = pq[i].p;
            q = pq[i].q;
            for(int j = 0; j < cntb; j++){
              int  k, m;
              k = km[j].k;
              m = km[j].m;
              if(p*m+q*k==b||p*k+q*m==b){
                  printf("YES\n");
                  flag = 1;
                  i = cnta; j = cntb;
              }
            }
        }
        if(flag==0) printf("NO\n");
    }
    return 0;
}

方法二：

判别式法：x = (-b+cnt_1)/2*a，即（2*a*x+(b-cnt_1)）（2*a*x+(b+cnt_1)）==0

下面问题就是 b-cnt_1 是否大于 0 ？因为cnt_1*cnt_1==b*b-4*a*c; a c不小于0 ，所以cnt_1 <= b 恒成立

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <cmath>
#include <iostream>
#define MAXN  1000000+10
#define ll long long
using namespace std;
int main(){
    ll t,a,b,c,cnt;
    scanf("%I64d",&t);
    while(t--){
        scanf("%I64d%I64d%I64d",&a,&b,&c);
        cnt = b*b - 4*a*c;
        if(cnt<0)  printf("NO\n");              //判别式小于0，方程无解
        else if(cnt==0)  printf("YES\n");       //判别式为0，方程有唯一解
        else{                                   
            int flag = 0;

            ll cnt_1 = sqrt(cnt);
            if(cnt_1*cnt_1==cnt) flag = 1;      //判别式可开方为整数

            if(flag == 0) printf("NO\n");
            else  printf("YES\n");

        }
    }
    return 0;
}