Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16394 Accepted Submission(s): 11552
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
Author
Ignatius.L
Recommend
就是将一个数 n 拆分成无序可以重复的数，可以几种方法。。
解题思路：
母函数。。。
(1+x+x^2+x^3+.....+x^n)*(1+x^2+x^4+....)*(1+x^3+x^6+...)*...
直接上代码：（可以当作模板）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e6+5;
const int mod = 1e9+7;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
LL gcd(LL a, LL b)
{
if(b == 0)
return a;
return gcd(b, a%b);
}
int c1[maxn];///每一次存的多项式中的数
int c2[maxn];///中间变量
int n;///拆分的数
/**
首先对c1初始化，由第一个表达式
(1+x+x^2+..x^n)初始化，
从0到n的所有x前面的系数都初始化为1.
**/
void Init()
{
for(int i=0; i<=n; i++)
{
c1[i] = 1;
c2[i] = 0;
}
}
int main()
{
while(cin>>n)
{
Init();
for(int i=2; i<=n; i++)///控制一共循环几次，也就是每个多项式乘几次
{
for(int j=0; j<=n; j++)///第一个表达式的系数
{
for(int k=0; k+j<=n; k+=i)///我们只需要 n 前面的系数
{
c2[k+j] += c1[j];
}
}
for(int j=0; j<=n; j++)///我们在用 c1[]当作第一个乘
{
c1[j] = c2[j];
c2[j] = 0;///c2不需要
}
}
cout<<c1[n]<<endl;
}
return 0;
}
