Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 100   Accepted Submission(s) : 83

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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 #include < iostream > #include < stdio.h > using namespace std; int main() { int i,k,j,n; int a[ 121 ],b[ 121 ]; while (scanf( " %d " , & n) != EOF) { for (i = 0 ;i <= n;i ++ ) { a[i] = 1 ; b[i] = 0 ; } for (i = 2 ;i <= n;i ++ ) { for (j = 0 ;j <= n;j ++ ) // (1+x+x2+x3..xn) for (k = 0 ;k + j <= n;k += i) // (1+x2+x4+x6..) 两层for循环　是用第一个循环中ｘ的ｊ次　来乘第二个循环中ｘ的ｋ次 b[j + k] += a[j]; for (j = 0 ;j <= n;j ++ ) { a[j] = b[j]; b[j] = 0 ; } } printf( " %d\n " ,a[n]); } return 0 ;}

转载于:https://www.cnblogs.com/cyiner/archive/2011/05/16/2048165.html