HDU 3903 Trigonometric Function(数学定理)
2016-03-09 09:08
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Trigonometric Function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 693 Accepted Submission(s): 277
Problem Description
Give you a triangle ABC. Get more information in the picture below.
Now, give you 6 integers a, b, c, n, m and k. a, b and c are triangle ABC`s three edges. Can you judge whether the result of the following fraction is rational number?
Input
There are several test cases in the input data.
Each case is just one line with 6 integers – a, b, c, n, m, k (0< a, b, c, n, m, k < 10^4) separated by spaces. The input data ensures that sin(kC) will not be equal with 0.
Output
Each case output “YES”, if the result of the fraction is rational number, otherwise “NO”.
Sample Input
2
1 1 1 1 1 1
3 4 5 6 7 7
Sample Output
NO
YES
若n为有理数,那么cos(n)一定是有理数
cos( k*n )一定是有理数
sin(n)确不一定。
所以只要判断分母是不是有理数就好了,sin(k*n)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 693 Accepted Submission(s): 277
Problem Description
Give you a triangle ABC. Get more information in the picture below.
Now, give you 6 integers a, b, c, n, m and k. a, b and c are triangle ABC`s three edges. Can you judge whether the result of the following fraction is rational number?
Input
There are several test cases in the input data.
Each case is just one line with 6 integers – a, b, c, n, m, k (0< a, b, c, n, m, k < 10^4) separated by spaces. The input data ensures that sin(kC) will not be equal with 0.
Output
Each case output “YES”, if the result of the fraction is rational number, otherwise “NO”.
Sample Input
2
1 1 1 1 1 1
3 4 5 6 7 7
Sample Output
NO
YES
若n为有理数,那么cos(n)一定是有理数
cos( k*n )一定是有理数
sin(n)确不一定。
所以只要判断分母是不是有理数就好了,sin(k*n)
#include <iostream> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #include <stdio.h> using namespace std; long long int a,b,c,n,m,k; int main() { int t; scanf("%d",&t); while(t--) { scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&n,&m,&k); long long int num=4*a*a*b*b-(a*a+b*b-c*c)*(a*a+b*b-c*c); long long int t=sqrt(1.0*num); if(t*t==num) printf("YES\n"); else printf("NO\n"); } return 0; }
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