• 【转】HDU1028


    转自博客园ID:2108,老卢同志

    http://www.cnblogs.com/--ZHIYUAN/p/6102893.html

    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:

    /*//整数拆分模板
    #include <iostream>
    using namespace std;
    const int lmax=10000;
    //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,
    //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。
    //用c1保存,然后在计算时用c2来保存变化的值。
    int c1[lmax+1],c2[lmax+1];
    int main()
    {
                int n, i, j, k ;
               // 计算的方法还是模拟手动运算,一个括号一个括号的计算,
               // 从前往后
               while ( cin>>n )
    
              {
                         //对于 1+x+x^2+x^3+ 他们所有的系数都是 1
                         // 而 c2全部被初始化为0是因为以后要用到 c2[i] += x ;
                         for ( i=0; i<=n; i++ )
    
                         {
                                    c1[i]=1;
                                    c2[i]=0;
                         }
                          //第一层循环是一共有 n 个小括号,而刚才已经算过一个了
                          //所以是从2 到 n
                         for (i=2; i<=n; i++)
    
                       {
                                     // 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
                                     //小括号里面的每一项计算。
                                    for ( j=0; j<=n; j++ )
                                     //第三层小括号是要控制每一项里面 X 增加的比例
                                     // 这就是为什么要用 k+= i ;
                                             for ( k=0; k+j<=n; k+=i )
    
                                            {
                                                     // 合并同类项,他们的系数要加在一起,所以是加法,呵呵。
                                                     // 刚开始看的时候就卡在这里了。
                                                     c2[ j+k] += c1[ j];
                                             }
                                   // 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。
                                  for ( j=0; j<=n; j++ )
    
                                  {
                                              c1[j] = c2[j] ;
                                              c2[j] = 0 ;
                                  }
                       }
                        cout<<c1[n]<<endl;
            }
             return 0;
    }
    #include<bitsstdc++.h>
    using namespace std;
    int c1[123],c2[123];
    void solve()
    {
        for(int i=0;i<=120;i++)
        {
            c1[i]=1;
            c2[i]=0;
        }
        for(int k=2;k<=120;k++)
        {
            for(int i=0;i<=120;i++)
            for(int j=0;j+i<=120;j+=k)
            c2[j+i]+=c1[i];
            for(int i=0;i<=120;i++)
            {
                c1[i]=c2[i];
                c2[i]=0;
            }
        }
    }
    int main()
    {
        int n;
        solve();
        while(scanf("%d",&n)!=EOF)
        {
            printf("%d
    ",c1[n]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liuzhanshan/p/6290451.html
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