华东交通大学2015年ACM“双基”程序设计竞赛1007

Problem G

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 681   Accepted Submission(s) : 192

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Problem Description

给定方程 x ^ 2 + bx + c = 0，求解方程由哪两个一次多项式(x + p) , (x + q)(p , q均为整数且p <= q)相乘得到

Input

输入包含多组测试样例(10组左右)，处理到文件结束，每组数据输入两个整数(b,c) (-100 <= b,c <= 100)

Output

每组测试数据中
如果存在输出两个整数p , q , 若不存在整数p,q,输出impossible;

Sample Input

3 2
1 2

Sample Output

1 2
impossible

Author

moonlike

 

首先当然要判断是否有解啦~(b*b-4*ac)是否小于0

然后用求根公式算两次，一次用int 一次用double 再算出误差，误差范围内就符合条件

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int main()
{
    int b,c;
    while(~scanf("%d%d",&b,&c))
    {
        int sum_1;
        double sum_2;
        //-b+-sqrt(b*b-4*a*c)/2
        if((b*b*1.0-4.0*c)<0)
        {
            printf("impossible
");
            continue;
        }
        sum_1=sqrt(b*b-4*c);
        sum_2=(double)sqrt(b*b-4*c);
        int ans_1,ans_2;
        double ans_3,ans_4;
        ans_1=(-1*b+sum_1)/2;
        ans_2=(-1*b-sum_1)/2;
        ans_3=(double)(-1.0*b+sum_2)/2;
        ans_4=(double)(-1.0*b-sum_2)/2;
        if(abs(ans_1-ans_3)<=1e-6&&abs(ans_2-ans_4)<=1e-6)
        {
            int x,y;
            x=-min(ans_1,ans_2);
            y=-max(ans_1,ans_2);
            printf("%d %d
",min(x,y),max(x,y));
        }
        else
        {
            printf("impossible
");
        }
    }
    return 0;
}