div2 574 D2. Submarine in the Rybinsk Sea (hard edition)

　　

一道维护贡献的好题   见代码

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s)
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define inf 0x3f3f3f3f
#define CLR(A,v)  memset(A,v,sizeof A)
typedef pair<int,int>pii;
//////////////////////////////////
const int N=1e5+10;
 
const ll mod=998244353;
ll val[N],x,p10[N],ans;
ll len[N],n;
ll cal(ll x)
{
    ll len=0;
    while(x)
    {
        val[len+1]+=x%10;
        len++;x/=10;
    }
    return len;
}
ll sol()
{
    cin>>n;
    rep(i,1,n)scanf("%lld",&x),len[i]=cal(x);
 
    p10[0]=1;
    rep(i,1,63)p10[i]=10*p10[i-1]%mod;
 
    repp(j,31,1)//枚举一个长度为的串  遍历1-n  判断其贡献
    rep(i,1,n)
    {
        if(len[i]>=j)
        ans=(ans+p10[(j-1)*2]*11%mod*val[j]%mod)%mod;
        else ans=(ans+p10[len[i]+j-1]*2%mod*val[j]%mod)%mod;
    }
    return ans;
}
 
int main()
{
    printf("%lld",sol());
}