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Right-angled Triangle
Accept: 52 Submit: 109
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
- A right triangle (or right-angled triangle, formerly called a rectangled triangle) has one 90° internal angle (a right angle). The side opposite to the right angle is the hypotenuse; it is the longest side in the right triangle. The other two sides are the legs or catheti (singular: cathetus) of the triangle. Right triangles conform to the Pythagorean Theorem, wherein the sum of the squares of the two legs is equal to the square of the hypotenuse, i.e., a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.
- An oblique triangle has no internal angle equal to 90°.
- An obtuse triangle is an oblique triangle with one internal angle larger than 90° (an obtuse angle).
- An acute triangle is an oblique triangle with internal angles all smaller than 90° (three acute angles). An equilateral triangle is an acute triangle, but not all acute triangles are equilateral triangles.
Input
Output
Sample Input
1240
Sample Output
15
Hint
There are five right-angled triangles where a + b + c ≤ 40. That are one right-angled triangle where a = 3, b = 4 and c = 5; one right-angled triangle where a = 6, b = 8 and c = 10; one right-angled triangle where a = 5, b = 12 and c = 13; one right-angled triangle where a = 9, b = 12 and c = 15; one right-angled triangle where a = 8, b = 15 and c = 17.
求满足以a,b为直角边，c为斜边，且满足a+b+c<=L的直角三角形的个数。
即本原毕达哥拉斯三元组数，枚举m，n，然后求出x，y，z乘以i倍，并且保证在范围内。
//187 ms 208KB
#include<stdio.h>
#include<math.h>
int gcd(int a,int b)
{return b==0?a:gcd(b,a%b);
}
void solve(int t)
{int tmp=sqrt(t),x,y,z,ans=0;for(int n=1; n<=tmp; n++)for(int m=n+1; m<=tmp; m++){if(2*m*m+2*m*n>t)break;//x+y+z>t结束if(n%2!=m%2){if(gcd(m,n)==1){x=m*m-n*n;y=2*m*n;z=m*m+n*n;for(int i=1;; i++){if(i*(x+y+z)>t)break;ans++;}}}}printf("%d\n",ans);
}
int main()
{int t;while(scanf("%d",&t)!=EOF)solve(t);return 0;
}
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