Ignatius and the Princess IIITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14352Accepted Submission(s): 10102
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
41020
Sample Output
542627
虽然母函数还不是太懂，，，但是拿着模板，还会刷出了一题。。
代码：#include <stdio.h>#define MAX 150int main(){int c1[MAX] , c2[MAX] , n;while(~scanf("%d",&n)){for(int i = 0 ; i <= n ; ++i){c1[i] = 1 ;c2[i] = 0 ;}for(int i = 2 ; i <= n ; ++i){for(int j = 0 ; j <= n ; ++j){for(int k = 0 ; j+k <= n ; k+=i){c2[j+k] += c1[j] ;}}for(int j = 0 ; j <= n ; ++j){c1[j] = c2[j] ;c2[j] = 0 ;}}printf("%d\n",c1[n]);}return 0 ;}与君共勉
学习会使你永远立于不败之地。
