丛林木马

假设$aa$字符串长度为$nn$,$bb$字符串长度为$mm\backslash \backslash$ 其实观察我们可以发现对于$aa$字符串中的每位都加了$mm$次,对于$bb$字符串的每位都加了$nn$次，然后注意每一位的值不是单一数字是需要*当前位的权值即可。$\backslash \backslash$ 小tips:最好用快速幂写$x^{y}x^y$,pow有可能不靠谱。

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<bits/stdc++.h>
# include<unordered_map>

# define eps 1e-9
# define fi first
# define se second
# define ll long long
# define int ll
# define x1 sb
# define y1 dsb
# define x2 ssb
# define y2 ddsb
// cout<<fixed<<setprecision(n) 
//bool operator<(const Node& x )
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int > PII; 
const int mod=998244353;
const int N=2e6+10;
const int Time=86400;
const int X=131;
const int inf=0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846; 
double e = exp(1);

int T,n,q,m,k,ans,res,sum,x1,y1,x2,y2;
map<int,int>mp;
string a,b;

int qsm(int a,int b){
	int res = 1;
	while(b){
		 if(b&1) res = res * a % mod;
		 a = a * a % mod;
		 b >>= 1;
	}
	return res % mod;
}
void solve(){
	     ans = 0;
	     cin >> a >> b;
	     n = a.size();m = b.size();
	     for(int i = 0 ; i < n ; i ++ ){
             int x = a[i]-'0';
			 x = x * qsm(10,n-i-1) % mod;
			 ans = (ans + x*m%mod) % mod;		 	
		 }
		 for(int i = 0 ; i < m ; i ++ ){
		    int x = b[i]-'0';
		    x = x * qsm(10,m-i-1) % mod;
		 	ans = (ans + x*n%mod) % mod;		 	
	}
	 cout<<ans<<"\n";
		
        
         
       			
		
}
/*
2 9 6 7 10
2 6 7 9 10
*/
signed main(){  
    std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 
    
     cin >> T;
     //T = 1;
    while(T--){
		solve();
	} 
    return 0; 
}