//计算x y之间距离
double distance(point a,point b){
return sqrt( (a.x - b.x)*(a.x - b.x) + (a.y -b.y)*(a.y -b.y) );
}
想要直接以这样的公式计算,但是当输入大于六位数会出错。需要输入小于等于1000000的数字。
所以有没有能够数据不溢出的解决方法。
当然计算结果只需要保留后两位即可。
参考代码:
#include <stdio.h>
#include <stdlib.h>
#include "math.h"
typedef struct point {
double x;
double y;
}point;
point array[30001];
double distance(point a,point b);
int main(int argc, char *argv[])
{
while(1){
int n,i,j,count;
double maxdis,temp;
scanf("%d",&n);
for(i = 0;i<n;i++){
scanf("%f %f",&array[i].x,&array[i].y);
}
for(i = 0;i<n;i++){
//printf("%d %d\n",array[i].x,array[i].y);
//printf("dis:%.2f\n",distance(array[0],array[i]));
}
maxdis = 0.0;
for(i = 0;i<n;i++){
for(j = i;j<n;j++){
if(distance((array[i]),(array[j])) > maxdis ){
//printf("maxdis = :%.2f\n",distance((array[i]),(array[j])));
maxdis =distance((array[i]),(array[j]));
}
}
}
printf("%.2f",maxdis);
}
system("pause");
return 0;
}
//计算x y之间距离
double distance(point a,point b){
return sqrt( (a.x - b.x)*(a.x - b.x) + (a.y -b.y)*(a.y -b.y) );
}
代码已经贴出。
PHP中文网2017-04-17 14:34:50
Please post the definition of point...
To be honest, if it is only 1 million W, double is easy and stress-free. Could it be that int is used...?
大家讲道理2017-04-17 14:34:50
I made the following modifications to your code:
Changed the scanf reading type from %f to %lf. The input data cannot be read using %f on my machine.
There is no problem compiling on my computer and the running results are normal.
The comment above has already solved this problem.
ringa_lee2017-04-17 14:34:50
Use high precision directly for very large numbers. This has nothing to do with whether the distance is calculated
天蓬老师2017-04-17 14:34:50
To tell you something unrelated to the problem... you can try using high precision.