1.将正整数n无序拆分成最大数为m的拆分方案个数,要求所有拆分方案不重复。
样例:
n = 5, m = 5,对应的拆分方案如下:
5 = 5
5 = 4 + 1
5 = 3 + 2
5 = 3 + 1 + 1
5 = 2 + 2 + 1
5 = 2 + 1 + 1 + 1
5 = 1 + 1 + 1 + 1 + 1
分析:
(1)当n=1,无论m为多少,只有{1}一种划分
(2)当m=1,无论n为多少,只有{1,1,1…}一种划分
(3)当n<m,f(n,m)=f(n,n)
(4)当n=m,如果划分中有n,则只有{n}一种划分;当划分中没有n,则f(n,n)=f(n,n-1)
f(n,n)= 1 + f(n,n - 1)
(5)当n>m,如果划分中有m,则{m,{x1,x2,…} = n - m},f(n,m)=f(n - m, m);当划分中没有m,f(n,m)=f(n,m - 1)
f(n,m) = f(n - m, m)+ f(n,m - 1)
import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); while (scanner.hasNext()){ int n = scanner.nextInt(); int m = scanner.nextInt(); int temp = integerhuafen(n, m); System.out.print(temp); } } public static int integerhuafen(int n, int m) { int dp[][] = new int[n + 1][m + 1]; for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++ ){ if(i == 1 || j == 1){ dp[i][j] = 1; }else if (i == j) { dp[i][j] = 1 + dp[i][j - 1]; }else if (i < j) { dp[i][j] = dp[i][i]; }else { dp[i][j] = dp[i - j][j] + dp[i][j - 1]; } } } return dp[n][m]; } }
2.将正整数n拆分成k份,每份不为空,不考虑顺序,求划分的种类。
样例:
n = 7, k = 3;
输出:
4
{1,1,5; 1,2,4; 1,3,3; 2,2,3}
分析:
(1)分的时候至少有一个1,相当于dp[n][k] = dp[n - 1][k - 1]
(2)分的时候没有1,dp[n][k] = dp[n - k][k]
=》 dp[n][k] = dp[n - 1][k - 1] + dp[n - k][k]
import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); while (scanner.hasNext()){ int n = scanner.nextInt(); int k = scanner.nextInt(); int [][] arr = new int [n + 1][k + 1]; arr[0][0] = 1; for (int i = 1; i <= n; i++){ for (int j = 1; j <= k; j++){ if(i >= j){ arr[i][j]=arr[i-j][j]+arr[i-1][j-1]; } } } System.out.println(arr[n][k]); } scanner.close(); } }