AtCoder Beginner Contest 132 F

数 sqrt 缩小范围

整除分块

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define ll long long

const int maxn=1e5+10;
const ll mod=1e9+7;
const double eps=1e-8;

ll f[110][maxn],add[maxn],cnt[maxn];

/**
大于sqrt(maxvalue)的x，
肯定是其它数到x，到从x到其它数

计数 用 整除分块
**/

int main()
{
    ll n,siz,mid,mmid,l,r,i,j,g=0,sum=0;
    scanf("%lld%lld",&siz,&n);
    mmid=sqrt(siz+eps);
    mid=siz/(mmid+1);
    g=0;
    for (l=1;l<=siz;l=r+1)
    {
        ///[l,r]
        r=siz/(siz/l);
//        printf("%lld %lld %lld
",l,r,siz/l);
        if (siz/l<=mid)
            cnt[siz/l]=r-l+1;
    }

    for (j=1;j<=mmid;j++)
        f[1][j]=1;
    for (i=2;i<=n;i++)
    {
        g=0;
        for (j=1;j<=mmid;j++)
            (g+=f[i-1][j])%=mod;
        for (j=1;j<=mmid;j++)
            f[i][j]=g;

        for (l=1;l<=mid;l++)
            add[l]=(add[l-1]+f[i-2][l])%mod;

        g=0;
        for (l=mid;l>=1;l--)
        {
            (g+=add[l]*cnt[l])%mod;
            (f[i][l]+=g)%=mod;
        }
    }


    for (j=1;j<=mmid;j++)
        (sum+=f[n][j])%mod;
    for (l=1;l<=mid;l++)
    {
        add[l]=(add[l-1]+f[n-1][l])%mod;
        (sum+=add[l]*cnt[l])%=mod;
    }

    ///
    memset(f,0,sizeof(f));
    g=0;
    for (l=mid;l>=1;l--)
    {
        (g+=cnt[l])%mod;
        f[1][l]=g;
    }
    n--;
    for (i=2;i<=n;i++)
    {
        g=0;
        for (j=1;j<=mmid;j++)
            (g+=f[i-1][j])%=mod;
        for (j=1;j<=mmid;j++)
            f[i][j]=g;

        for (l=1;l<=mid;l++)
            add[l]=(add[l-1]+f[i-2][l])%mod;

        g=0;
        for (l=mid;l>=1;l--)
        {
            (g+=add[l]*cnt[l])%mod;
            (f[i][l]+=g)%=mod;
        }
    }

    for (j=1;j<=mmid;j++)
        (sum+=f[n][j])%mod;
    for (l=1;l<=mid;l++)
    {
        add[l]=(add[l-1]+f[n-1][l])%mod;
        (sum+=add[l]*cnt[l])%=mod;
    }


    printf("%lld",sum);
    return 0;
}
/*
special
100=10*10

100 3
1 1 100
2 2 50
3 3 33
4 4 25
5 5 20
6 6 16
7 7 14
8 8 12
9 9 11
10 10 10
///
11 11 9
12 12 8
13 14 7
15 16 6
17 20 5
21 25 4
26 33 3
34 50 2
51 100 1


23 3
1 1 23
2 2 11
3 3 7
4 4 5
///
5 5 4
6 7 3
8 11 2
12 23 1


31622*log(n) *100

10 5
1 1 10
2 2 5
3 3 3

4 5 2
6 10 1

*/