Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12521 Accepted Submission(s): 8838
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
Sample Output
整数拆分无限取，跟着包子做的题，就当做模板来用吧。
#include <stdio.h>
#define maxn 122
int c1[maxn], c2[maxn];
int main()
{
int n, i, j, k;
while(scanf("%d", &n) != EOF){
for(i = 0; i <= n; ++i){
c1[i] = 1; c2[i] = 0;
}
for(i = 2; i <= n; ++i){
for(j = 0; j <= n; ++j)
for(k = j; k <= n; k += i)
c2[k] += c1[j];
for(k = 0; k <= n; ++k){
c1[k] = c2[k]; c2[k] = 0;
}
}
printf("%d\n", c1[n]);
}
return 0;
}
HDU1028 Ignatius and the Princess III 【母函数模板题】,布布扣,bubuko.com
HDU1028 Ignatius and the Princess III 【母函数模板题】
原文：http://blog.csdn.net/chang_mu/article/details/38123083
