hdu1028 整數劃分
阿新 • • 發佈：2019-02-20
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21096 Accepted Submission(s): 14720
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!"
Output For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
import java.util.Scanner;
public class hdu1028 {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan=new Scanner(System.in);
int n;
int record[][]=new int[120+5][120+5];
while(scan.hasNextInt()){
n=scan.nextInt();
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==1||j==1){
record[i][j]=1;
}else if(i<j){
record[i][j]=record[i][i];
}else if(i==j){
record[i][j]=1+record[i][j-1];
}else if(i>j){
record[i][j]=record[i-j][j]+record[i][j-1];
}
}
}
System.out.println(record[n][n]);
}
}
}