[算法 – Java的]分享宽松利用返回
需要线程列出可能的情况下,当N量分成一硬币面值[在].
import java.util.Scanner;
class java_chiatien
{
int n, N, a[], b[], sum=0;
public void input()
{
Scanner in = new Scanner(System.in);
System.out.print("Nhap so tien can chia: ");
N = in.nextInt();
System.out.print("Nhap so loai tien le: ");
n = in.nextInt();
a = new int [n];
b = new int [N];
for (int i=0; i<n; i++)
{
System.out.printf("Nhap loai tien thu %d : ",i+1);
a[i] = in.nextInt();
}
in.close();
}
public void chiatien(int i)
{
for (int k = 0; k < n; k++)
{
b[i] = a[k];
if (i==0 || (i >0 && b[i] >= b[i-1]))
{
sum += b[i];
if (sum <= N)
{
if (sum == N)
{
for (int l=0; l<=i; l++)
System.out.print(b[l]+ " ");
System.out.print("n");
}
else
chiatien(i+1);
}
sum -= b[i];
}
}
}
public static void main(String[] agrs)
{
java_chiatien ct = new java_chiatien();
ct.input();
System.out.print("nCac cach chia tien :n");
ct.chiatien(0);
}
}



另一种为类似文章:
让我们来数N的方式分析 ( ñ<=100000 ) 正整数的总和.
注意 2 唯一的差别是如何的术语的次序被认为是相同的. 例 4 有 5 经过分析:
4 = 1 + 1 + 1 + 1
4 = 1 + 1 + 2
4 = 1 + 3
4 = 2 + 2
4 = 4
双向分析 4 = 1 + 3 = 3 + 1 只计算一次.
由于结果将是非常大的,所以你只需要做出设法的除法的余 1000000000 ( 10^9 ).
输入
自然数N只.
产量
在所述的方法找到了10 ^ 9的数目的剩余部分.
例
输入:
4
产量:
5
uses crt; var a: array[1..10000] of 0..9; n, i, j: longint; procedure print(vitri: integer); begin write(n,' = '); for j:= 1 to vitri do if j= vitri then write(a[j]) else write(a[j],' + '); writeln; end; begin clrscr; write('Nhap n: '); readln(n); for i:= 1 to n do a[i]:= 1; print(i); i:= n; while a[1] &lt;&gt; n do begin a[i-1]:= a[i-1] + a[i]; dec(i); print(i); if i&gt;1 then while (a[i]-a[i-1] &gt;= 2) do begin dec(a[i]); inc(a[i-1]); print(i); end; end; readln end.从参考代码,你 tranminhchien 在 codevn.org
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