HDU 1028Ignatius and the Princess III(母函数简单题)

 Ignatius and the Princess III

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

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

 题意理解：问给出的那个数能有多少种不同的数相加

 #include<iostream>
 #include<stdio.h>
 using namespace std;
 int main() {
     int N;
     int c1[],c2[];
     while(cin>>N) {
         int i,j,k;
         for(i=; i<=N; i++) { //初始化第一个表达式的系数
             c1[i]=;
             c2[i]=;
         }
         for(i=; i<=N; i++) {
             //从第二个表达式开始，因为有无限制个，所以有n个表达式
             for(j=; j<=N; j++) {
                 //从累乘的表达式后的一个表达式第一个到最后一个
                 for(k=; k+j<=N; k+=i) {
                     //k为第j个变量的指数，第i个表达式每次累加i
                     c2[j+k]+=c1[j];
                 }
             }
             for(j=; j<=N; j++) {
                 //滚动数组算完一个表达式后更新一次
                 c1[j]=c2[j];
                 c2[j]=;
             }
         }
         printf("%d\n",c1[N]);
     }
     return ;
 }