Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17917 Accepted Submission(s): 12558
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!"
"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
看了题解,学了两种方法。一种是母函数,一种是dp。
母函数:组合数学方法,第一次接触。
此题构造的母函数(1+x^1+x^2+x^3...+x^n)(1+x^2+x^4+x^6...+x^2n).....
第一项表示(0个1,1个1,2个1,3个1...),第二项表示(0个2,1个2,2个2,3个2,4个2...)以此类推。
展开后,每一项的指数表示划分的这个数,系数表示该数的划分数。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
import java.util.*; import java.io.*; public class Main { public static int cal(int n) { int c1[]=new int [n+1]; int c2[]=new int [n+1]; for(int i=0;i<=n;i++) { c1[i]=1; c2[i]=0; } for(int i=2;i<=n;i++) { for(int j=0;j<=n;j++) for(int k=0;k+j<=n;k+=i) c2[j+k]+=c1[j]; for(int j=0;j<=n;j++) { c1[j]=c2[j]; c2[j]=0; } } return c1[n]; } public static void main(String[] args) { Scanner in=new Scanner(System.in); int n; while(in.hasNext()) { n=in.nextInt(); System.out.println(cal(n)); } } }
dp:
dp[i][j]表示i这个数划分为最大加数不超过j的划分数。
if(i>j) dp[i][j]=dp[i][j-1]+dp[i-j][j];
else if(i==j) dp[i][j]=dp[i][j-1]+1;
else if(i<j) dp[i][j]=dp[i][i];
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; int dp[150][150]; int main() { int n,m; //dp[1][1]=1; for(int i=1; i<=150; i++) for(int j=1; j<=150; j++) { if(j==1) dp[i][j]=1; else if(i==j) dp[i][j]=dp[i][j-1]+1; else if(i>j) dp[i][j]=dp[i][j-1]+dp[i-j][j]; else if(i<j) dp[i][j]=dp[i][i]; } while(scanf("%d",&n)!=EOF) { printf("%d ",dp[n][n]); } return 0; }