Ignatius and the Princess III 1028 (母函数)
2015-09-08 20:30
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Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15971 Accepted Submission(s): 11266
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//母函数公式为:g(x)=(1+x+x^2+x^3+x^4+......)*(1+x^2+x^4+x^6+......)*(1+x^3+x^6+x^9+......)*......
#include<stdio.h> int a[1010],b[1010]; int main(){ int n,i,j,k; while(scanf("%d",&n)!=EOF) { for(i=0;i<=n;i++) { a[i]=1; b[i]=0; } for(i=2;i<=n;i++) { for(j=0;j<=n;j++) { for(k=0;k+j<=n;k+=i) { b[k+j]+=a[j]; } } for(j=0;j<=n;j++) { a[j]=b[j]; b[j]=0; } } printf("%d\n",a ); } return 0; }
//拓展一下,下面的这个是母函数公式为g(x)=(1+x)(1+x^2)(1+x^3)......(不能保证是否正确,因为没有相应的题)
#include<stdio.h> int a[1010],b[1010]; int main(){ int n,i,j,k,t; while(scanf("%d",&n)!=EOF) { for(i=0;i<=n;i++) { a[i]=1; b[i]=0; } t=3; for(i=3;i<=n;i++) { for(j=0;j<=t;j++) { for(k=0;k+j<=n;k+=i) { b[k+j]+=a[j]; } } t+=i; for(j=0;j<=n;j++) { a[j]=b[j]; b[j]=0; } } printf("%d\n",a ); } return 0; }
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