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!"
"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[125],b[125]; // 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 = 0;i <= 125;i ++) { a[i] = 0; b[i] = 1; } for(i = 2;i <= 125;i ++) { for(j = 0;j <= 125;j ++) { for(k = 0;k+j <= 125; 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 = 0;j <= 125;j ++) { b[j] = a[j]; a[j] = 0; } } while(~scanf("%d",&n)) printf("%d ",b[n]);
return 0;
}