HDU 1028 Ignatius and the Princess III(生成函数)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24796    Accepted Submission(s): 17138

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

 

Author

Ignatius.L

 

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生成函数的裸题

对于每个数构造一个多项式$ax^i$表示权值为$i$的方案数为$a$

初始时$a$为1

然后全乘起来就行

 

#include<cstdio>
#include<cstring>
#include<algorithm>
const int MAXN = 121;
inline int read() { 
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int cur[MAXN], nxt[MAXN];
main() {
    int N;
    while(scanf("%d", &N) != EOF) {
        memset(cur, 0, sizeof(cur));
        for(int i = 0; i <= N; i++) cur[i] = 1;
        for(int i = 2; i <= N; i++) {
            for(int j = 0; j <= N; j++) 
                for(int k = 0; j + i * k <= N; k++) 
                    nxt[j + k * i] += cur[j];
            memcpy(cur, nxt, sizeof(nxt));
            memset(nxt, 0, sizeof(nxt));
        }
        printf("%d
", cur[N]);
    }
    return 0;
}