1 条题解
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0
C :
#include<stdio.h> int main() { int n; double x1,x2,x3,y1,y2,y3; while(scanf("%d",&n)!=EOF,n) { while(n--) { scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3); printf("%.1lf %.1lf\n",(x1+x2+x3)/3,(y1+y2+y3)/3); } } return 0; }
C++ :
#include<stdio.h> int main() { int n; double x1,x2,x3,y1,y2,y3; while(scanf("%d",&n)!=EOF,n) { while(n--) { scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3); printf("%.1lf %.1lf\n",(x1+x2+x3)/3,(y1+y2+y3)/3); } } return 0; }
Java :
import java.util.*; public class Main{ public static void main(String[] args){ Scanner in=new Scanner(System.in); while(in.hasNextInt()){ int n=in.nextInt(); if(n==0) break; double[] x=new double[3]; double[] y=new double[3]; for(int i=1;i<=n;i++){ double ave1=0,ave2=0; for(int j=0;j<3;j++){ x[j]=in.nextDouble(); y[j]=in.nextDouble(); } for(int k=0;k<3;k++){ ave1+=x[k]; ave2+=y[k]; } ave1=ave1/3.0; ave2=ave2/3.0; System.out.printf("%.1f",ave1); System.out.print(" "); System.out.printf("%.1f\n",ave2); } } } }
- 1
信息
- ID
- 1664
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者