Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12758 Accepted Submission(s): 9016
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
/*题解：
母函数问题， 这里可视为母函数模板.
*/
#include<cstdio>
#include<cstring>
int main()
{
int i,j,k,n,c1[1010],c2[1010];
while(scanf("%d",&n)!=EOF)
{
memset(c2,0,sizeof(c2));
for(i=0; i<=n; i++)
c1[i]=1; //在未乘时，每个未知数前的系数为1
for(i=2; i<=n; i++) //括号数
{
for(j=0; j<=n; j++) //j和k均为相乘时双方的指数
{
for(k=0; j+k<=n; k+=i)
{
c2[j+k]+=c1[j]; //c1,c2存储的是系数
}
}
for(j=0; j<=n; j++)
{
c1[j]=c2[j]; //c2只是临时储存，在这里释放，将c2值赋给c1;
c2[j]=0;
}
}
printf("%d\n",c1[n]);
}
return 0;
}