容斥原理

C - Ubiquity

https://atcoder.jp/contests/abc178/tasks/abc178_c

全部-没有0的-没有9的+重复减去的值（既没有0又没有9）

#include<bits/stdc++.h>
#define int long long 
#define pb push_back
#define fi first
#define se second 
#define bg begin()
#define ed end()
#define rbg rbegin()
#define all(x) x.bg,x.ed
#define cy cout<<"YES"<<endl
#define cn cout<<"NO"<<endl
#define de(x) cout<<x<<"###"<<endl
using namespace std;
const long long inf=2e18;
typedef long long ll;
typedef pair<ll,ll>pii;
typedef vector<ll>vi;
const double eps=1e-10;
const int mod=1e9+7;
const int N=2e5+5;
ll ksm(int a,int b){
    int res=1;
    while(b){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
void solve(){
    int n;
    cin>>n;
    int ans=(ksm(10,n)-2*ksm(9,n)+ksm(8,n))%mod;
    ans=(ans+mod)%mod;
    cout<<ans<<endl;
}        
signed main(){
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif 
    ios::sync_with_stdio(false);
      cin.tie(0);
    //int t;
    //cin>>t;
    //while(t--){
        solve();
    //}
    return 0;
}