夹角有多大II

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11539    Accepted Submission(s): 5983

Problem Description

这次xhd面临的问题是这样的：在一个平面内有两个点，求两个点分别和原点的连线的夹角的大小。

注：夹角的范围[0，180]，两个点不会在圆心出现。

 

Input

输入数据的第一行是一个数据T，表示有T组数据。
每组数据有四个实数x1,y1,x2,y2分别表示两个点的坐标,这些实数的范围是[-10000,10000]。

 

Output

对于每组输入数据，输出夹角的大小精确到小数点后两位。

 

Sample Input

21 1 2 21 1 1 0

 

Sample Output

0.0045.00

 

Author

xhd

 

Source

ACM程序设计期末考试_热身赛(感谢 xhd & 8600)

 

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#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include <set>#include<algorithm>#include <limits.h>#include <ctype.h>#include <map>#include <stack>#include <sstream>typedef long long LL;using namespace std;const int N = 105;int data[N][N];int dp[N][N];char c[100][100] = {' '};#define PI 3.1415926int main() {    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T;    cin >> T;    double x1, y1, x2, y2;    while(T--){        //if(n == 0 && m== 0) break;        scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);        double a = sqrt(x1 * x1 + y1 * y1);        double b = sqrt(x2 * x2 + y2 * y2);        double c = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));        double cos = (a * a + b * b - c * c) / (2 * a * b);        double res = acos(cos) * 180 / PI;        if(res > 180){            res -= 180;        }        printf("%.2lf\n", res);    }    return 0;}