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  • Growling Gears

    http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11587

    G Growling Gears
    The Best Acceleration Production Company specializes in multi-gear engines. The performance
    of an engine in a certain gear, measured in the amount of torque produced, is not constant:
    the amount of torque depends on the RPM of the engine. This relationship can be described
    using a torque-RPM curve.
    RPM
    Torque
    Gear 1
    Gear 2
    The torque-RPM curve of the gears given in the second sample input.
    The second gear can produce the highest torque.
    For the latest line of engines, the torque-RPM curve of all gears in the engine is a parabola
    of the form T = −aR2 + bR + c, where R is the RPM of the engine, and T is the resulting
    torque.
    Given the parabolas describing all gears in an engine, determine the gear in which the
    highest torque is produced. The first gear is gear 1, the second gear is gear 2, etc. There will
    be only one gear that produces the highest torque: all test cases are such that the maximum
    torque is at least 1 higher than the maximum torque in all the other gears.
    Input
    On the first line one positive number: the number of test cases, at most 100. After that per test
    case:
    • one line with a single integer n (1 ≤ n ≤ 10): the number of gears in the engine.
    • n lines, each with three space-separated integers a, b and c (1 ≤ a, b, c ≤ 10 000): the
    parameters of the parabola T = −aR2 +bR+c describing the torque-RPM curve of each
    engine.
    Output
    Per test case:
    • one line with a single integer: the gear in which the maximum torque is generated.14 Problem G: Growling Gears
    Sample in- and output
    Input Output
    3
    1
    1 4 2
    2
    3 126 1400
    2 152 208
    2
    3 127 1400
    2 154 208
    1
    2
    2

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <malloc.h>
    #define Max(a,b) (a>b?a:b)
    #define Min(a,b) (a<b?a:b)
    #define MAX 999999999
    #define LL long long
    #define M 6666666
    using namespace std;
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
         int a,b,c,n,i,s[15];
          scanf("%d",&n);
           for(i=1;i<=n;i++)
           {
               scanf("%d%d%d",&a,&b,&c);
               s[i]=(4*(-1*a)*c-b*b)/(4*((-1)*a));
            }
            int max=0,g;
            for(i=1;i<=n;i++)
            {
               //printf("s=%d
    ",s[i]);
                if(s[i]>max)
                    {
                     g=i;
                     max=s[i];
                    }
            }
            printf("%d
    ",g);
    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cancangood/p/4725399.html
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