Problem Description “Well, it seems the first problem is too easy. I will let you know how foolish you are later.” feng5166 says.
“The second problem is, given an positive integer N, we define an equation like this: N=a[1] a[2] a[3] … a[m]; a[i]>0,1<=m<=N; My question is how many different equations you can find for a given N. For example, assume N is 4, we can find: 4 = 4; 4 = 3 1; 4 = 2 2; 4 = 2 1 1; 4 = 1 1 1 1; so the result is 5 when N is 4. Note that “4 = 3 1” and “4 = 1 3” is the same in this problem. Now, you do it!”
Input The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.
Output For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
Sample Input 4 10 20
Sample Output 5 42 627
思路： （i，j）（i>=j）代表的含义是i为n，j为划分的最大的数字。 边界：a(i,0) = a(i, 1) = a(0, i) = a(1, i) = 1; i|j==0时，无论如何划分，结果为1；
当（i>=j）时， 划分为{j,{x1,x2…xi}},{x1,x2,…xi}的和为i-j, {x1,x2,…xi}可能再次出现j，所以是(i-j)的j划分，所以划分个数为a(i-j,j); 划分个数还需要加上a(i,j-1)（累加前面的）;
当（i < j）时， a[i][j]就等于a[i][i]；
代码语言：javascript复制import java.util.Scanner;
public class Main{
static int a[][] = new int[125][125];
public static void main(String[] args) {
dabiao();
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n = sc.nextInt();
System.out.println(a[n][n]);
}
}
private static void dabiao() {
for(int i=0;i<121;i ){
a[i][0]=1;
a[i][1]=1;
a[0][i]=1;
a[1][i]=1;
}
for(int i=2;i<121;i ){
for(int j=2;j<121;j ){
if(j<=i){
a[i][j]=a[i][j-1] a[i-j][j];
}else{
a[i][j]=a[i][i];
}
}
}
}
}
