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P5735 【深基7.例1】距离函数

G13460488251 @ 2022-09-29 19:41:10

# include <stdio.h>
# include <math.h>

int main(void)
{
    int i;
    float x1,x2,x3,sum;
    int a[3];
    int b[3];
    for (i = 0;i < 3;++i)
    {
        scanf("%d %d",&a[i],&b[i]);
    }
        x1 = sqrt((a[0]-a[1])*(a[0]-a[1])+(b[0]-b[1])*(b[0]-b[1]));
        x2 = sqrt((a[0]-a[2])*(a[0]-a[2])+(b[0]-b[2])*(b[0]-b[2]));
        x3 = sqrt((a[1]-a[2])*(a[1]-a[2])+(b[1]-b[2])*(b[1]-b[2]));
        sum = x1 + x2 + x3;
        printf("%.2f\n",sum);
        return 0;
}

by Ctrl_Z @ 2022-09-29 19:42:38

@G13460488251 您可以下载数据


by wf715 @ 2022-09-29 19:49:07

输入的坐标是实数,不一定是整数


by Wy_x @ 2022-09-29 19:55:15

  1. float a[3]; float b[3];

  2. scanf("%f %f",&a[i],&b[i]);


by Wy_x @ 2022-09-29 19:57:05

更改后 AC code

# include <stdio.h>
# include <math.h>

int main(void)
{
    int i;
    float x1,x2,x3,sum;
    float a[3];//改正后
    float b[3];//改正后
    for (i = 0;i < 3;++i)
    {
        scanf("%f %f",&a[i],&b[i]);
        //改正后
    }
        x1 = sqrt((a[0]-a[1])*(a[0]-a[1])+(b[0]-b[1])*(b[0]-b[1]));
        x2 = sqrt((a[0]-a[2])*(a[0]-a[2])+(b[0]-b[2])*(b[0]-b[2]));
        x3 = sqrt((a[1]-a[2])*(a[1]-a[2])+(b[1]-b[2])*(b[1]-b[2]));
        sum = x1 + x2 + x3;
        printf("%.2f\n",sum);
        return 0;
}

by Wy_x @ 2022-09-29 19:57:32

@G13460488251


by G13460488251 @ 2022-10-02 18:36:49

谢谢大佬


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