第一题
pdd01.png
import java.util.Scanner;/**
* @Author: Taoyongpan
* @Date: Created in 15:10 2018/4/3
*/public class Test05 { public static void main(String[] args){
Scanner sc = new Scanner(System.in); while (sc.hasNext()){ int n = sc.nextInt(); int k = sc.nextInt(); int[] arr = new int[105]; int a = 0; int b = 0; for (int i = 0 ; i< n;i++){
a = sc.nextInt();
b = sc.nextInt();
a+=50;
b+=50; for (int j = a; j <= b;j++){
arr[j]++;
}
} int max = -1; int min = 105; for (int i = 0 ; i<101;i++){ if (arr[i]>=k){ if (i>=max)
max = i; if (i<=min)
min = i;
}
}
max-=50;
min-=50; if (min<max&&min>100&&max<0)
System.out.println(min+" "+max); else
System.out.println("error");
}
}
}第二题
pdd02.png
解法一，不可取，有点暴力
import java.text.DecimalFormat;import java.util.Scanner;/**
* @Author: Taoyongpan
* @Date: Created in 15:10 2018/4/3
*/public class Test06 {
public static void main(String[] args){
Scanner sc = new Scanner(System.in); while (sc.hasNext()){
String str = sc.nextLine(); char[] c = str.toCharArray(); double h = 0; int m = 0; if(c.length==4){
m = (c[2]-'0')*10+(c[3]-'0');
h = ((c[0]-'0')+((m*1.0)/60));
}else if (c.length ==5){
m = (c[3]-'0')*10+(c[4]-'0');
h = (((c[0]-'0')*10+(c[1]-'0'))+((m*1.0)/60));
} if (m<=59 && m>=0 && h<24 && h>=0){ if ((Math.abs((h%12)*5-m)*6)%1 == 0){ if (Math.abs((h%12)*5-m)*6<=180)
System.out.println((int)(Math.abs((h%12)*5-m)*6)); else
System.out.println((int)(360-Math.abs((h%12)*5-m)*6));
}else {
DecimalFormat df = new DecimalFormat("#.0"); if (Math.abs((h%12)*5-m)*6<=180)
System.out.println(df.format(Math.abs((h%12)*5-m)*6)); else
System.out.println(df.format(360-Math.abs((h%12)*5-m)*6));
}
}else {
System.out.println("error");
}
}
}
}解法二
import java.text.DecimalFormat;import java.util.Scanner;/**
* @Author: Taoyongpan
* @Date: Created in 20:35 2018/4/3
*/public class Test09 { public static void main(String[] args){
Scanner sc = new Scanner(System.in); while (sc.hasNext()){
String str = sc.nextLine();
String[] s = str.split(":"); int h, m;
h = Integer.valueOf(s[0]);
m = Integer.valueOf(s[1]); double gap = (30 * h - 5.5 * m) % 360; if (gap>180){
gap = 360-gap;
} if (gap%1==0){
System.out.println((int)gap);
}else {
DecimalFormat df = new DecimalFormat("#.0");
System.out.println(df.format(gap));
}
}
}
}第三题
pdd03.png
#include<bits/stdc++.h> using namespace std;
typedef long long ll;
const double pi=acos(-1.0);
const double eps=1e-8;struct point{
double x,y;
point(double a=0,double b=0){
x=a,y=b;
}
}p[110];
double xmult(point p1,point p2,point p0){ return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}int dblcmp(double x){ if(x<-eps) return -1; if(x>eps) return 1; return 0;
}int dot_inline(point a,point b,point c){ return dblcmp(xmult(a,b,c));
}int main(){ int n; cin>>n; for(int i=0;i<n;i++){ cin>>p[i].x>>p[i].y;
}
ll cnt=0; for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ for(int k=j+1;k<n;k++){ if(dot_inline(p[i],p[j],p[k])) cnt++;
}
}
} cout<<cnt<<endl; return 0;
}这题的代码是我室友写出来的，我的思路是，求斜率，当三个点在一条直线上的时候不能构成，其他都行，暴力遍历一遍就能 算出结果；
作者：Taoyongpan
链接：https://www.jianshu.com/p/c40aa3836b77
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