【JZOJ2700】数字

description

---

analysis

• 把$D(n)$打表出来可以发现从$1$开始$1$到$9$循环

• 容易知道$D(x)=(x-1)mod9+1$

• 其实这里有个$trick$，由于$lcm(1...10)=22680$

• 所以如果一个数$n$是喜欢的数，那么$n+22680$也是喜欢的数

• 那么把前$22680$个喜欢的数的表打出来，做前缀和，后面的答案可以直接得到

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define O3 __attribute__((optimize("-O3")))

using namespace std;

ll a[22685],pre[22685];
bool bz[22685];
ll n;

O3 inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
O3 inline ll f(ll x)
{
	return (x/22680)*pre[22680]+pre[x%22680];
}
O3 int main()
{
	//freopen("T1.in","r",stdin);
	fo(i,1,22680)
	{
		a[i]=(i-1)%9+1;
		if (i*a[i]<=22680)bz[i*a[i]]=1;
	}
	fo(i,1,22680)pre[i]=pre[i-1]+bz[i];
	n=read();
	while (n--)
	{
		ll x=read(),y=read();
		printf("%lld\n",f(y)-f(x-1));
	}
	return 0;
}