Ignatius and the Princess III

时间:2024-06-23 10:05:08

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10312    Accepted Submission(s): 7318

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
思路:一开始拿到这个题目以为是找规律,有递推关系什么的,最后找了好长时间没找到规律,上网查了一下才发现是用母函数做,就是把数的加法和指数乘法的幂的加法联系起来,母函数:G(x)=(1+x+x^2+x^3+x^4+.....)*(1+x^2+x^4+x^6+....)*(1+x^3+x^6+x^9+....)*... ,x^n的系数就是n的拆分方案数!其实这个不难理解,因为x^n的系数是多少就表明有多少个x^n相加得来,换句话说就是有多少种x的幂之和的拼凑方案,即本题所求。
#include<stdio.h>
int a[],b[]; // a[i]表示x^i的系数,为临时值,b[i]表示x^i的系数,为最终值;
int
main()
{

int
i,j,k,n;
for
(i =;i <=;i ++)
{

a[i] =;
b[i] =;
}

for
(i =;i <=;i ++)
{

for
(j =;j <=;j ++)
{

for
(k =;k+j <=; k += i)
a[k+j] += b[j]; //因为x^(k+j)是从x^j得来的,故它的系数应该在原有系数的数值的基础上加上x^j                 
                      的系数(这是关键的重点!!!这就是为什么我们要用两个数组的目的)
}

for
(j =;j <=;j ++)
{

b[j] = a[j];
a[j] =;
}
}

while
(~scanf("%d",&n))
printf("%d\n",b[n]);
  return 0;
}