HDU1028 Ignatius and the Princess III【整数划分+记忆化递归】

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26672 Accepted Submission(s): 18371

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

问题链接：HDU1028 Ignatius and the Princess III
问题描述：（略）
问题分析：
    这个问题是整数划分问题，与参考链接中的问题属于同一种问题。这里给出2种题解，基于参考链接的题解代码更加简洁。
    本题采用记忆化递归的解法。
    这个题也可以用母函数来解，是一个母函数的裸题。
程序说明：
    本题与参考链接是同一题，使用参考链接的程序提交就AC了。
    需要注意类型，以防止数据溢出！
参考链接：POJ1664 放苹果【递推+记忆化递归】
题记：朋友交多了，容易遇见熟人。

AC的C语言程序如下：

/* HDU1028 Ignatius and the Princess III */

#include 
#include 

#define N 160
long long a[N + 1][N + 1];

long long ways(int m, int n)
{
    if(a[m][n])
        return a[m][n];
    else {
        if(m == 0 || n == 1)
            return a[m][n] = 1;
        else if(n > m)
            return a[m][n] = ways(m, m);
        else
            return a[m][n] = ways(m - n, n) + ways(m, n - 1);
    }
}

int main(void)
{
    memset(a, 0, sizeof(a));

    int n;
    while(~scanf("%d", &n))
        printf("%lld\n", ways(n, n));

    return 0;
}

AC的C语言程序如下：

/* HDU1028 Ignatius and the Princess III */

#include 
#include 

#define N 160
long long c[N + 1][N + 1];

long long ways(int m, int n)
{
    if(c[m][n] >= 0)
        return c[m][n];
    else {
        if(m == 0)
            return c[m][n] = 1;
        else if(n == 0)
            return c[m][n] = 0;
        else if(n <= m) {
            if(c[m - n][n] < 0)
                c[m - n][n] = ways(m - n, n);
            if(c[m][n - 1] < 0)
                c[m][n - 1] = ways(m, n - 1);
            return c[m][n] = c[m - n][n] + c[m][n - 1];
        } else
            return c[m][n - 1] >= 0 ? c[m][n - 1] : (c[m][n - 1] = ways(m, n - 1));
    }
}

int main(void)
{
    memset(c, 0xFF, sizeof(c));

    int n;
    while(~scanf("%d", &n)) {
        printf("%lld\n", ways(n, n));
    }

    return 0;
}