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  • UVa 10566

    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %lluDescription

    A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?

    Input

    Input starts with an integer T (≤ 10), denoting the number of test cases.

    Each test case contains three positive floating point numbers giving the values of xy, and c.

    Output

    For each case, output the case number and the width of the street in feet. Errors less than 10-6 will be ignored.

    Sample Input

    4

    30 40 10

    12.619429 8.163332 3

    10 10 3

    10 10 1

    Sample Output

    Case 1: 26.0328775442

    Case 2: 6.99999923

    Case 3: 8

    Case 4: 9.797958971


    列出方程组, 利用二分求解精确值。

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    
    double x, y, c;
    double ans(double z) {
        return 1 - c/sqrt(x*x - z*z) - c/sqrt(y*y - z*z);
    }
    int main() {
        int t;
        scanf("%d", &t);
        int cnt = 0;
        while(t--){
            scanf("%lf%lf%lf",&x, &y, &c);
            double l = 0, mid, r = min(x, y);
            while(r - l > 1e-8) {
                mid = (l + r)/2;
                if(ans(mid) > 0) l = mid;
                else  r = mid;
            }
            printf("Case %d: %.8lf
    ", ++cnt, mid);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770902.html
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