Topic links: HDU 1028
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
Solution
The meaning of problems
Given \ (n-\) , find \ (n-\) of the division number.
Thinking
Ordinary generating function. Generating function \ (G (x) = ( 1 + x + x ^ 2 + ...) (1 + x ^ 2 + x ^ 4 + ...) (1 + x ^ 3 + x ^ 6 + .. .) ... \) .
\ ((1 + x + x ^ 2 + ...) = (x ^ {0 \ times1} + x ^ {1 \ times1} + x ^ {2 \ times1} + ...) \) representative of digital not \ (1 \) , with a digital \ (1 \) , with two digital \ (1 \) ......
Dynamic programming version, see here .
Code
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200;
int c1[maxn], c2[maxn];
void init() {
for(int i = 0; i < maxn; ++i) {
c1[i] = 1;
c2[i] = 0;
}
for(int i = 2; i < maxn; ++i) {
for(int j = 0; j < maxn; ++j) {
for(int k = 0; k + j < maxn; k += i) {
c2[k + j] += c1[j];
}
}
for(int j = 0; j < maxn; ++j) {
c1[j] = c2[j];
c2[j] = 0;
}
}
}
int main() {
init();
int n;
while(~scanf("%d", &n)) {
printf("%d\n", c1[n]);
}
return 0;
}