hdu 1028 对某个数n的m划分数

末蓝、 2022-08-03 15:56 128阅读 0赞

# Ignatius and the Princess III #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 15687    Accepted Submission(s): 11063**

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

Recommend

We have carefully selected several similar problems for you:   [1171][]  [1398][]  [2152][]  [1709][]  [1059][]

//hdu 1028
    //dp[i][j]代表j的小于i划分的总数，则dp[i][j]=dp[i-1][j]+dp[i][j-i];
    //dp[i-1][j]肯定是前面划分的数的总和，dp[i][j-i]则是从第i-1个划分到第i个划分的增量;
    //如果这i个划分全部大于0，他们的和肯定是j，对于每个数，把它减去1，奇迹的得到了j-i的i划分;
    //因为他们有i个，每个都减去1，自然总共减去了i，减去后发现是j-i的i划分，所以增量就是这个。
    #include<bitset>
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;
        char c;
        while((c=getchar())<'0' || c>'9');
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res;
    }
    ll dp[122][122];
    int main()
    {
        dp[0][0]=1;
        for(int i=1;i<=120;i++) //外层循环是划分个数
        {
            for(int j=0;j<=120;j++) //内层循环是对于数j的划分
            {
                if(j-i>=0)
                {
                    dp[i][j]=dp[i-1][j]+dp[i][j-i];
                }
                else dp[i][j]=dp[i-1][j];
            }
        }
        //for(int i=2;i<=20;i++)cout<<dp[i][i]<<" ";
        int n;
        while(~scanf("%d",&n))
        {
            printf("%I64d\n",dp[n][n]);
        }
        return 0;
    }

[1171]: http://acm.hdu.edu.cn/showproblem.php?pid=1171
[1398]: http://acm.hdu.edu.cn/showproblem.php?pid=1398
[2152]: http://acm.hdu.edu.cn/showproblem.php?pid=2152
[1709]: http://acm.hdu.edu.cn/showproblem.php?pid=1709
[1059]: http://acm.hdu.edu.cn/showproblem.php?pid=1059