正题

luogu题目链接:https://www.luogu.org/problemnew/show/P4193

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题目大意

定义一个函数$D(x)$和$S(x)$，$S(x)$表示$x$的各位之和
$$D(n)=\{\begin{array}{c}\\ D(S(n)),S\ge 10\\ S(n)\\ \end{array}$$
求$L\sim R$之间有多个$x$满足$x=D(k)\ast k$

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解题思路

因为$(n-S(n))mod9=0$,所以$D(n)=(n-1)mod9+1$
$$22680modx=0(x\in [1..9])$$
若一个数$n=D(k)\ast k$，那么$n+22680=(k+\frac{22680}{D(k)})\ast D(k)$
证明:

$$(k+\frac{22680}{D(k)})\ast D(k)-k\ast D(k)=22680$$
$$(k+\frac{22680}{D(k)}-k)\ast D(k)=22680$$
$$(k+\frac{22680}{D(k)}-k)\ast D(k)=22680$$
$$k\ast D(k)+22680-k\ast D(k)=22680$$
$$22680=22680$$
证毕

然后之间根据循环节预处理$1\sim 22680$的就好了

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$code$

#include<cstdio>
#define LCM 22680
#define ll long long
using namespace std;
ll n,f[1000000],ans;
ll D(ll x)
{return (x-1)%9+1;}
ll ask(ll x)//1~x的个数
{return x/LCM*ans+f[x%LCM];}
int main()
{scanf("%lld",&n);for(ll i=1;i<=LCM;i++)//预处理{for(ll j=1;j<=9;j++)if(D(i/j)==j&&i%j==0){f[i]=1;ans++;break;}f[i]+=f[i-1];}while(n--){ll l,r;scanf("%lld%lld",&l,&r);printf("%lld\n",ask(r)-ask(l-1));}
}