Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16028 Accepted Submission(s): 11302
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
题解：相当于取砝码，每个砝码可以取多次；
4=1+1+1+1=1+1+2=1+3=2+2；
则母函数为：
相当于1的砝码取法从0....n,2的砝码0....n/2。。。。。。k的砝码从0....n/k；
代码：
#include<stdio.h>
const int MAXN=;
int main(){
int a[MAXN],b[MAXN],N;
while(~scanf("%d",&N)){
int i,j,k;
for(i=;i<=N;i++){
a[i]=;b[i]=;
}
for(i=;i<=N;i++){
for(j=;j<=N;j++)
for(k=;k+j<=N;k+=i)
b[j+k]+=a[j];
for(j=;j<=N;j++)
a[j]=b[j],b[j]=;
}
printf("%d\n",a[N]);
}
return ;
}
extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
typedef unsigned long long ULL;
void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(char *x){scanf("%s",x);}
}
const int MAXN = ;
int a[MAXN],b[MAXN];
int main(){
int N;
while(~scanf("%d",&N)){
for(int i = ;i <= N;i++)a[i] = ,b[i] = ;
for(int i = ;i <= N;i++){
for(int j = ;j <= N;j++){
for(int k = ;j + k <= N;k += i){
b[j + k] += a[j];
}
}
for(int j = ;j <= N;j++)a[j] = b[j],b[j] = ;
}
printf("%d\n",a[N]);
}
return ;
}