Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19589    Accepted Submission(s): 13709

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

题意：

拆数共有多少总方案

代码：

 1 /*//整数拆分模板
 2 #include <iostream>
 3 using namespace std;
 4 const int lmax=10000;
 5 //c1是用来存放展开式的系数的，而c2则是用来计算时保存的，
 6 //他是用下标来控制每一项的位置，比如 c2[3] 就是 x^3 的系数。
 7 //用c1保存，然后在计算时用c2来保存变化的值。
 8 int c1[lmax+1],c2[lmax+1];
 9 int main()
10 {
11             int n, i, j, k ;
12            // 计算的方法还是模拟手动运算，一个括号一个括号的计算，
13            // 从前往后
14            while ( cin>>n )
15
16           {
17                      //对于 1+x+x^2+x^3+ 他们所有的系数都是 1
18                      // 而 c2全部被初始化为0是因为以后要用到 c2[i] += x ;
19                      for ( i=0; i<=n; i++ )
20
21                      {
22                                 c1[i]=1;
23                                 c2[i]=0;
24                      }
25                       //第一层循环是一共有 n 个小括号，而刚才已经算过一个了
26                       //所以是从2 到 n
27                      for (i=2; i<=n; i++)
28
29                    {
30                                  // 第二层循环是把每一个小括号里面的每一项，都要与前一个
31                                  //小括号里面的每一项计算。
32                                 for ( j=0; j<=n; j++ )
33                                  //第三层小括号是要控制每一项里面 X 增加的比例
34                                  // 这就是为什么要用 k+= i ;
35                                          for ( k=0; k+j<=n; k+=i )
36
37                                         {
38                                                  // 合并同类项，他们的系数要加在一起，所以是加法，呵呵。
39                                                  // 刚开始看的时候就卡在这里了。
40                                                  c2[ j+k] += c1[ j];
41                                          }
42                                // 刷新一下数据，继续下一次计算，就是下一个括号里面的每一项。
43                               for ( j=0; j<=n; j++ )
44
45                               {
46                                           c1[j] = c2[j] ;
47                                           c2[j] = 0 ;
48                               }
49                    }
50                     cout<<c1[n]<<endl;
51         }
52          return 0;
53 }

 1 #include<bits\stdc++.h>
 2 using namespace std;
 3 int c1[123],c2[123];
 4 void solve()
 5 {
 6     for(int i=0;i<=120;i++)
 7     {
 8         c1[i]=1;
 9         c2[i]=0;
10     }
11     for(int k=2;k<=120;k++)
12     {
13         for(int i=0;i<=120;i++)
14         for(int j=0;j+i<=120;j+=k)
15         c2[j+i]+=c1[i];
16         for(int i=0;i<=120;i++)
17         {
18             c1[i]=c2[i];
19             c2[i]=0;
20         }
21     }
22 }
23 int main()
24 {
25     int n;
26     solve();
27     while(scanf("%d",&n)!=EOF)
28     {
29         printf("%d\n",c1[n]);
30     }
31     return 0;
32 }