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package leo.class22;
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/**
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* @author Leo
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* @ClassName splitNumber
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* @DATE 2021/1/14 10:02 下午
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* @Description
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*
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* 给定一个正数1，裂开的方法有一种，
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* (1) 给定一个正数2，裂开的方法有两种，(1和1)、
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* (2) 给定一个正数3，裂开的方法有三种，(1、1、1)、(1、2)、
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* (3) 给定一个正数4，裂开的方法有五种，(1、1、1、1)、(1、1、2)、(1、3)、(2、2)、 (4)
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* 给定一个正数n，求裂开的方法数。 动态规划优化状态依赖的技巧
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*/
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public class splitNumber {
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static class Recursion {
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public static int number(int n) {
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if (n <= 0) {
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return 0;
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}
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return process(n, 1);
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}
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private static int process(int rest, int pre) {
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if (rest == 0) {
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return 1;
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}
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if (rest <= pre) {
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return rest - pre == 0 ? 1 : 0;
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}
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int ans = 0;
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for (int j = pre; j <= rest; j++) {
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ans += process(rest - j, j);
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}
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return ans;
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}
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}
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static class Dp{
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public static int number(int n) {
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if (n <= 0) {
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return 0;
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}
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int[][] dp = new int[n + 1][n + 1];
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for (int i = 1; i <= n; i++) {
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dp[i][i] = 1;
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dp[i][0] = 1;
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}
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for (int pre = n - 1; pre > 0; pre--) {
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for (int rest = pre + 1; rest <= n; rest++) {
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int ans = 0;
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for (int j = pre; j <= rest; j++) {
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ans+= dp[j][rest - j];
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}
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dp[pre][rest] = ans;
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}
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}
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return dp[1][n];
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}
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}
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static class OptDp{
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public static int number(int n) {
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if (n <= 0) {
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return 0;
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}
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int[][] dp = new int[n + 1][n + 1];
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for (int i = 1; i <= n; i++) {
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dp[i][i] = 1;
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dp[i][0] = 1;
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}
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for (int pre = n - 1; pre > 0; pre--) {
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for (int rest = pre + 1; rest <= n; rest++) {
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//下面的格子
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dp[pre][rest] = dp[pre+1][rest];
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//左侧的格子
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dp[pre][rest] += dp[pre][rest - pre];
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}
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}
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return dp[1][n];
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}
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}
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public static void main(String[] args){
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System.out.println( Recursion.number(13));
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System.out.println(Dp.number(13));
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System.out.println(OptDp.number(13));
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}
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}
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