Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24967 Accepted Submission(s): 17245

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

分析：

整数的划分问题

关于这个整数划分请参考我的这篇博客

https://www.cnblogs.com/yinbiao/p/8672198.html

关于这个题目

有两种做法

1.暴力打表

打表就不是折磨你自己了，而是折磨电脑了

运行程序之后大概要等几分钟才有结果

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 125
/*LL f(int n,int m)
{
    if(n==1||m==1)
        return 1;
    else if(n==m&&n>1)
    {
        return f(n,n-1)+1;
    }else if(n<m)
    {
        return f(n,n);
    }else if(n>m)
    {
        return f(n,m-1)+f(n-m,m);
    }
}*/
int main()
{
    LL a[]={0,1,2,3,5,7,11,15,22,30,42,56,77,101,135,176,231,297,385,490,627,792,1002,1255,1575,1958,2436,3010,3718,4565,5604,6842,8349,10143,12310,14883,17977,21637,26015,31185,37338,44583,53174,63261,75175,89134,105558,124754,147273,173525,204226,239943,281589,329931,386155,451276,526823,614154,715220,831820,966467,1121505,1300156,1505499,1741630,2012558,2323520,2679689,3087735,3554345,4087968,4697205,5392783,6185689,7089500,8118264,9289091,10619863,12132164,13848650,15796476,18004327,20506255,23338469,26543660,30167357,34262962,38887673,44108109,49995925,56634173,64112359,72533807,82010177,92669720,104651419,118114304,133230930,150198136,169229875,190569292,214481126,241265379,271248950,304801365,342325709,384276336,431149389,483502844,541946240,607163746,679903203,761002156,851376628,952050665,1064144451,1188908248,1327710076,1482074143,1653668665,1844349560,2056148051,2291320912,2552338241,2841940500};
    /*for(int i=1;i<max_v;i++)
    {
        printf("%I64d,",f(i,i));
    }
    printf("\n**\n");*/
    int n;
    while(~scanf("%d",&n))
    {
        printf("%I64d\n",a[n]);
       // printf("%I64d\n",f(n,n));
    }
    return 0;
}

2.另外一种做法

记忆化搜索

就是在搜索之前先判断一下，搜过了就不再搜索了

跟dp一个意思

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define max_v 125
int dp[max_v][max_v];
LL f(int n,int m)
{
    if(dp[n][m]!=-1)//判断有没有搜过
        return dp[n][m];
    if(n==1||m==1)
        return dp[n][m]=1;
    else if(n==m&&n>1)
    {
        return dp[n][m]=f(n,n-1)+1;
    }else if(n<m)
    {
        return dp[n][m]=f(n,n);
    }else if(n>m)
    {
        return dp[n][m]=f(n,m-1)+f(n-m,m);
    }
}
int main()
{
    //两种做法
    //LL a[]={0,1,2,3,5,7,11,15,22,30,42,56,77,101,135,176,231,297,385,490,627,792,1002,1255,1575,1958,2436,3010,3718,4565,5604,6842,8349,10143,12310,14883,17977,21637,26015,31185,37338,44583,53174,63261,75175,89134,105558,124754,147273,173525,204226,239943,281589,329931,386155,451276,526823,614154,715220,831820,966467,1121505,1300156,1505499,1741630,2012558,2323520,2679689,3087735,3554345,4087968,4697205,5392783,6185689,7089500,8118264,9289091,10619863,12132164,13848650,15796476,18004327,20506255,23338469,26543660,30167357,34262962,38887673,44108109,49995925,56634173,64112359,72533807,82010177,92669720,104651419,118114304,133230930,150198136,169229875,190569292,214481126,241265379,271248950,304801365,342325709,384276336,431149389,483502844,541946240,607163746,679903203,761002156,851376628,952050665,1064144451,1188908248,1327710076,1482074143,1653668665,1844349560,2056148051,2291320912,2552338241,2841940500};
    /*for(int i=1;i<max_v;i++)
    {
        printf("%I64d,",f(i,i));
    }
    printf("\n**\n");*/
    memset(dp,-1,sizeof(dp));//没有搜索的标记
    int n;
    while(~scanf("%d",&n))
    {
        //printf("%I64d\n",a[n]);
        printf("%I64d\n",f(n,n));
    }
    return 0;
}

HDU 1028 Ignatius and the Princess III 整数的划分问题（打表或者记忆化搜索）

Ignatius and the Princess III

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