P1028 数的计算

递推做法：

我们先列出前几项，找规律

0	1	2	3	4	5	6	7	8	9	10	11	12	13	14
1	1	2	2	4	4	6	6	10	10	14	14	20	20	26

于是乎我们惊奇的发现

f [ 1 ] = 1

f [ 2 ] = f [ 1 ] + 1

f [ 3 ] = f [ 1 ] + 1

f [ 4 ] = f [ 1 ] + f [ 2 ] + 1

f [ 5 ] = f [ 1 ] + f [ 2 ] + 1

f [ 6 ] = f [ 1 ] + f [ 2 ] + f [ 3 ] + 1

也就是 f [ n ] = ∑ f [ i ] ( i = 1 , 2 , 3 .... n / 2 ) + 1

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>

using namespace std;

int ans[1005],n;

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=i/2;j++)
          ans[i]+=ans[j];
        ans[i]++;
    }
      
    printf("%d",ans[n]);
    
    return 0;    
    
}

DP做法：

我们先列出前几项，找规律

0	1	2	3	4	5	6	7	8	9	10	11	12	13	14
1	1	2	2	4	4	6	6	10	10	14	14	20	20	26

于是乎我们又一次惊奇的发现

1. f [ 1 ] = f [ 0 ] f [ 3 ] = f [ 2 ] f [ 5 ] = f [ 4 ]

即 if ( i % 2 == 1 ) f [ i ] = f [ i - 1 ]

2. f [ 2 ] = f 1 + f 0 f [ 4 ] = f 3 + f 2 f [ 6 ] = f 5 + f 3 f [ 8 ] = f 7 + f 4

即 f [ i ] = f [ i - 1 ] + f [ i / 2 ] ( i % 2 == 0 )

若（i % 2 == 1） f [ i ] = f [ i - 1 ]

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>

using namespace std;

int f[1005],n;

int main()
{
    scanf("%d",&n);
    
    f[0]=1;f[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(i%2==1) f[i]=f[i-1-1]+f[(i-1)/2];
        else f[i]=f[i-1]+f[i/2];
    }
      
    printf("%d",f[n]);
    
    return 0;    
    
}

P1028 数的计算

P1028 数的计算

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