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

?

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#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\n");             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\n",min(x,y),max(x,y));         }         else         {             printf("impossible\n");         }     }     return 0; }