HDU-1028 Ignatius and the Princess III（DP[完全背包]||生成函数）

此处有目录↑

Ignatius and the Princess III (跳转)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

解法一：DP[完全背包]

母函数入门题，但是看着这种母函数的模版就觉得像完全背包，实际一看题，第一反应还是完全背包

因为本题数的拆分不考虑顺序，所以只考虑方案的序列非减即可，用完全背包的方法从数1开始枚举到数n即可

#include 
#include 
#include 

using namespace std;

int n;
int dp[125];//dp[i]表示凑成i的方案数

int main() {
    while(1==scanf("%d",&n)) {
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=1;i<=n;++i) {
            for(int j=i;j<=n;++j) {//从前往后是完全背包，从后往前是01背包
                dp[j]+=dp[j-i];
            }
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}

解法二：生成函数

了解到生成函数就是抽象成表达式然后进行化简，但是写的时候还是忍不住的想成DP，看来还需要多做题以养成意识

#include 
#include 
#include 

using namespace std;

int n;
int a[125],t[125];//a[i]表示x^i的系数，即凑成数i的方案数

int main() {
    while(1==scanf("%d",&n)) {
        for(int i=0;i<=n;++i) {//初始化第一个表达式 (1+x+x^2+…+x^n)，系数均为1
            a[i]=1;
            t[i]=0;
        }
        for(int i=2;i<=n;++i) {//遍历第i个表达式 (1+x^i+x^(2*i)+…)
            for(int j=0;j<=n;++j) {//遍历前i-1个表达式展开的表达式的x^j的系数
                for(int k=0;j+k<=n;k+=i) {//遍历第i个表达式的x^k的系数【此处完全可以写成for(int k=j;k<=n;k+=i) {t[k]+=a[j];}，则k表示展开的表达式x^k，但不便于理解表达式的展开】
                    t[j+k]+=a[j];
                }
            }
            for(int j=0;j<=n;++j) {//得到前i个表达式展开的系数
                a[j]=t[j];
                t[j]=0;
            }
        }
        printf("%d\n",a[n]);
    }
    return 0;
}