[C++知识库]abc238 C(偷偷用__int128做题是吧)

题目
题意: 求Sigma f(i) , f(x): <=x且和x位数相同的数的个数。 n = 1e18
思路: 猜!
1-9
10-99
100-999
都是从1开始，等差数列求和就完了。
当时用了一个笨方法，都爆ll了，改用__int128偷偷过了。
实际上是取模没有加括号，导致爆了，学废了。
时间复杂度: O(logn)
代码:

// Problem: C - digitnum
// Contest: AtCoder - Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）
// URL: https://atcoder.jp/contests/abc238/tasks/abc238_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 998244353;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
#define int __int128
int n,m,k,T;
int pow(int a,int k)
{
	int res = 1;
	for(int i=0;i<k;++i) res = res * a;
	return res; 
}
int qpow(int a,int k)
{
	int res = 1;
	while(k)
	{
		if(k&1) res = res * a % mod;
		a = a * a % mod;
		k >>= 1;
	}
	return res;
}
int ni(int x)
{
	return qpow(x,mod-2);
}
int count(int x)
{
	int res = 0;
	while(x)
	{
		res ++ ;
		x /= 10;
	}
	return res;
}
int cal(int x)
{
	return (x - pow(10,count(x)-1) + 1);
}
int fun(int x)
{
	int cnt = count(x) - 1;
	int res = 0;
	for(int i = 9;cnt--;i = i * 10)
	{
		int tmp = i * ni(2) % mod  * (i+1) %mod;
		res = (res + tmp) % mod;
	}
	return res;
}
void solve()
{
   ll x; read(x);
   ll ans = fun(x);
   ll le = pow(10,count(x)-1);
   x = x - le + 1;
   ll tmp =  x * ni(2) % mod  * (x+1) % mod;
   ans = (ans + tmp) % mod;
   ans = (ans + mod) % mod;
   write(ans);
}
signed main(void)
{  
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}
// 100000000000000000
// 900000000000000000
// 999999999999999999

// Problem: C - digitnum
// Contest: AtCoder - Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）
// URL: https://atcoder.jp/contests/abc238/tasks/abc238_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 998244353;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
#define int long long
int n,m,k,T;
int pow(int a,int k)
{
	int res = 1;
	for(int i=0;i<k;++i) res = res * a;
	return res; 
}
int qpow(int a,int k)
{
	int res = 1;
	while(k)
	{
		if(k&1) res = res * a % mod;
		a = a * a % mod;
		k >>= 1;
	}
	return res;
}
int ni(int x)
{
	return qpow(x,mod-2);
}
int count(int x)
{
	int res = 0;
	while(x)
	{
		res ++ ;
		x /= 10;
	}
	return res;
}
int cal(int x)
{
	return (x - pow(10,count(x)-1) + 1);
}
int fun(int x)
{
	int cnt = count(x) - 1;
	int res = 0;
	for(int i = 9;cnt--;i = i * 10)
	{
		res = (res + (i % mod) * ((i+1)%mod ) % mod * ni(2) %mod )%mod;
	}
	return res;
}
void solve()
{
   ll x; read(x);
   ll ans = fun(x);
   ll le = pow(10,count(x)-1);
   x = x - le + 1;
   ans = (ans + (x % mod) * ((x+1)%mod) % mod * ni(2) %mod ) % mod;
   ans = (ans + mod) % mod;
   write(ans);
}
signed main(void)
{  
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}
// 100000000000000000
// 900000000000000000
// 999999999999999999