# Decimal point (Euler's theorem)

Meaning: Idea:
f ( i ) f(i) f(i) the conclusion is that if i i If i contains only 2 and 5, then it is 0, otherwise it is m a x ( m c ( 2 ) , m c ( 5 ) ) + 1 max(mc(2),mc(5))+1 max(mc(2),mc(5))+1, m c mc mc stands for the power of i. why? Let's prove it roughly
Suppose the length of the current loop section is n, and then the shape starting from the first bit, such as
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AC Code:

```//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 400010;
const int mod=998244353;
{
int res=0;
int f=1;
char c=getchar();
while(c>'9' ||c<'0')
{
if(c=='-')	f=-1;
c=getchar();
}
while(c>='0'&&c<='9')
{
res=(res<<3)+(res<<1)+c-'0';
}
return res;
}
#define int long long
int solve(int x){
int now=1;
int res=0;
for(int i=0;i<=60;i++){
int nn=now;
int cost=i;
if(nn>x)
break;
for(int j=0;;j++){
cost=max(i,j)+1;
int cnt=x/nn;

res+=max(0ll,(cnt-cnt/2-cnt/5+cnt/10-1))%mod*cost%mod;
res%=mod;
nn*=5;
if(nn>x)
break;
}
now*=2;
}
//	cout<<res<<endl;
return res;
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int q;
cin>>q;
while(q--){
int l,r;
cin>>l>>r;
cout<<(solve(r)-solve(l-1)+mod)%mod<<endl;
}
return 0;

}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/

```

Posted on Fri, 01 Oct 2021 16:02:56 -0400 by explore