Codeforces - 1195D1 - Submarine in the Rybinsk Sea (easy edition) - 水题

https://codeforc.es/contest/1195/problem/D1

给\(n\)个等长的十进制数串，定义操作\(f(x,y)\)的结果是“从\(y\)的末尾开始一个一个交替放得到的数”，求\(\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}f(a_i,a_j)\)

很明显每个第\(k\)位（从1开始）都会恰好在\(2k\)和\(2k-1\)位各贡献\(n\)次。预处理出\(pow10\)然后暴力就可以了。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll a[100005];

const ll mod=998244353;

ll ans;

ll calc(ll x){

    ll res=0;

    ll base=1;

    while(x){

        ll r=x%10ll;

        res=(res+base*11ll%mod*r%mod)%mod;

        x/=10;

        base=(base*100ll)%mod;

    }

    //cout<<res<<endl;

    return res%mod;

}

int main() {

#ifdef Yinku

    freopen("Yinku.in", "r", stdin);

    //freopen("Yinku.out", "w", stdout);

#endif // Yinku

    int n;

    while(~scanf("%d", &n)) {

        for(int i=1;i<=n;++i){

            scanf("%lld",&a[i]);

        }

        ans=0;

        for(int i=1;i<=n;i++){

            ans=(ans+calc(a[i]))%mod;

        }

        printf("%lld\n",ans*n%mod);

    }

}