hdu 1028 Ignatius and the Princess III
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9729 Accepted Submission(s): 6876
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
Author
Ignatius.L
母函数模板题
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 int c1[1000],c2[1000]; 3 int main() 4 { 5 int n,i,j,k,c; 6 while(scanf("%d",&n)!=EOF) 7 { 8 for(i=0;i<=n;i++) 9 { 10 c1[i]=0; 11 c2[i]=1; 12 } 13 for(i=1;i<=n;i++) 14 { 15 for(j=0;j<=n;j++) 16 { 17 for(k=0;k+j<=n;k+=i) 18 { 19 c2[j+k]+=c1[j]; 20 } 21 } 22 for(c=0;c<=n;c++) 23 { 24 c1[c]=c2[c]; 25 c2[c]=0; 26 } 27 } 28 printf("%d\n",c1[n]); 29 30 } 31 return 0; 32 }