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
41020
Sample Output
542627 解析： 母函数入门题，首先母函数
G(x) = (1 + x^1 + x^2..+x^n)(1 + x^2 + x^4 + x^6 + ...)(1 + x^3 + x^6 +..)(..)(1 + x^n) 其中(1 + x^1 + x^2..+x^n)中x的指数为1出现的次数 x^3 ==》 x^(1*3) 其余同理，比如第二个表达式 (1 + x^2 + x^4 + x^6 + ...) x^2 ==> x^(2*1) x^6 ==> x^(2*3)
乘法原理的应用：每一个表达式 表示的都是 某个变量的所有取值 【比如第一个表达式 表示'1'可以取的值(即n拆分后'1'出现的次数)可以为 {0，1，2...n}】每个变量的所有取值的乘积 就是问题的所有的解（在本问题中表现为‘和’）例子：4 = 2 + 1 + 1就是 x^(1 * 2)【'1'出现2次】 * x^(2 * 1)【'2'出现1次】 * x^(3 * 0)【'3'出现0次】 * x^(4 * 0)【..】 的结果 所以 G(x)展开后 其中x^n的系数就是 n的拆分解个数
# includeint main(){ int C1[123], C2[123], n; while(scanf("%d", &n) != EOF) { for(int i = 0; i <= n; i++)//初始化 第一个表达式 目前所有指数项的系数都为1 { C1[i] = 1; C2[i] = 0; } for(int i = 2; i <= n; i++)//第2至第n个表达式 { for(int j = 0; j <= n; j++)//C1为前i-1个表达式累乘后各个指数项的系数 { for(int k = 0; j + k <= n; k += i)//k为第i个表达式每个项的指数 第一项为1【即x^(i * 0)】（指数k=0），第二项为x^(i * 1)（指数为k=i）， 第三项为x^(i * 2)... 所以k的步长为i { C2[j + k] += C1[j];//(ax^j)*(x^k) = ax^(j+k) -> C2[j+k] += a 【第i个表达式每一项的系数都为1； a为C1[j]的值（x^j的系数）； C2为乘上第i个表达式后各指数项的系数】 } } for(int j = 0; j <= n; j++)//刷新当前累乘结果各指数项的系数 { C1[j] = C2[j]; C2[j] = 0; } } printf("%d\n",C1[n]); } return 0;}
出处http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=21943&messageid=2&deep=1