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.
Sample Input
3 1 4 4 0 0 1 1 3 1
Sample Output
YES YES NO
Hint
Source
“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)
题意:给你一个逻辑公式,给出a,b,c,让你判断这个逻辑公式对错;
思路:
只要有x在a*x*x+b*x+c==0这个式子成立,那么x一定是整数,即:
p q p->q
0 0 1
0 1 1
1 0 0
1 1 1
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main(){ int T; scanf("%d",&T); while(T--){ double a,b,c; scanf("%lf%lf%lf",&a,&b,&c); if(a==0&&b==0){ if(c==0) printf("NO\n"); //当a=b=c=0时,所有的x都符合,所以是NO else printf("YES\n");//c!=0,前式为0,无论后式为何值,结果都为0 } else if(a==0){ if((-c/b)==(int)(-c/b)) printf("YES\n"); else printf("NO\n"); } else if(b*b-4*a*c>=0){ int flag=0; for(int i=-5;i<=5;i++) if(a*i*i+b*i+c==0) flag++; if(b*b-4*a*c==0&&flag==1) printf("YES\n"); else if(b*b-4*a*c>0&&flag==2) printf("YES\n"); else printf("NO\n"); } else printf("YES\n"); } return 0; }