计算直线的交点数
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9357 Accepted Submission(s): 4226
Problem Description
平面上有n条直线,且无三线共点,问这些直线能有多少种不同交点数。 比如,如果n=2,则可能的交点数量为0(平行)或者1(不平行)。
Input
输入数据包含多个测试实例,每个测试实例占一行,每行包含一个正整数n(n<=20),n表示直线的数量.
Output
每个测试实例对应一行输出,从小到大列出所有相交方案,其中每个数为可能的交点数,每行的整数之间用一个空格隔开。
Sample Input
2
3
Sample Output
0 1
0 2 3
题解:用st存一下i条直线的点数目的种类;然后枚举平行线的条数j;当前点的个数就是*iter+平行的*不平行的,画个图看下就好了;
想了下,写下就过了;有点dp的意思
代码:
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<set> using namespace std; set<int>st[25]; set<int>::iterator iter; void db(){ st[1].insert(0); for(int i = 2; i < 25; i++){ st[i].insert(0); for(int j = 1; j < i; j++){ for(iter = st[i - j].begin(); iter != st[i - j].end(); iter++){ int dot = *iter + (i - j) * j; st[i].insert(dot); } } } } int main(){ int n; db(); while(~scanf("%d", &n)){ for(iter = st[n].begin(); iter != st[n].end(); iter++){ if(iter != st[n].begin())printf(" "); printf("%d", *iter); } puts(""); } return 0; }
java:
package com.lanqiao.week1; import java.util.Arrays; import java.util.HashSet; import java.util.Iterator; import java.util.Scanner; import java.util.Set; public class hdu1466 { private static Scanner cin = null; static Set<MyInteger>[] set = null; static MyInteger[][] arr = new MyInteger[25][]; private static final int MAXN = 25; static{ cin = new Scanner(System.in); set = new HashSet[MAXN]; } static class MyInteger implements Comparable{ int v; @Override public int hashCode() { return v; } public int getV() { return v; } public MyInteger(int v) { super(); this.v = v; } @Override public String toString() { return v + ""; } @Override public boolean equals(Object obj) { MyInteger t = (MyInteger)obj; return t.v == this.v; } @Override public int compareTo(Object o) { MyInteger t = (MyInteger)o; return this.v - t.getV(); } } private static void getSet(){ for(int i = 0; i < MAXN; i++){ set[i] = new HashSet<MyInteger>(){}; } set[1].add(new MyInteger(0)); for(int i = 2; i < MAXN; i++){ set[i].add(new MyInteger(0)); for(int j = 1; j < i; j++){ Iterator<MyInteger> iter = set[i - j].iterator(); while(iter.hasNext()){ //System.out.println("zz"); set[i].add(new MyInteger(iter.next().getV() + j * (i - j))); } } } for(int i = 1; i < MAXN; i++){ arr[i] = new MyInteger[set[i].size()]; set[i].toArray(arr[i]); Arrays.sort(arr[i]); } } public static void main(String[] args) { getSet(); while(cin.hasNext()){ int N = cin.nextInt(); for(int i = 0; i < arr[N].length; i++){ if(i != 0){ System.out.print(" "); } System.out.print(arr[N][i].getV()); }System.out.println(); } } }