solution

$O(n^3)$ で愚直に全部試す。 手を抜いて $O(n^3 \log n)$ だとおそらく落ちる。

note

ところで 30% rule 厳しすぎではないか。 空行以外のどの行を消してもコンパイル通らないはずなのに怒られるの理不尽すぎる。

implementation

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
using namespace std;
class DigitRotation { public: int sumRotations(string x); };

template <int32_t MOD>
struct mint {
int64_t data;  // faster than int32_t a little
mint() = default;  // data is not initialized
mint(int64_t value) : data(value) {}  // assume value is in proper range
inline mint<MOD> operator + (mint<MOD> other) const { int64_t c = this->data + other.data; return mint<MOD>(c >= MOD ? c - MOD : c); }
inline mint<MOD> operator - (mint<MOD> other) const { int64_t c = this->data - other.data; return mint<MOD>(c <    0 ? c + MOD : c); }
inline mint<MOD> operator * (mint<MOD> other) const { int64_t c = this->data * int64_t(other.data) % MOD; return mint<MOD>(c < 0 ? c + MOD : c); }
inline mint<MOD> & operator += (mint<MOD> other) { this->data += other.data; if (this->data >= MOD) this->data -= MOD; return *this; }
inline mint<MOD> & operator *= (mint<MOD> other) { this->data = this->data * int64_t(other.data) % MOD; if (this->data < 0) this->data += MOD; return *this; }
mint<MOD> pow(uint64_t k) const {
mint<MOD> x = *this;
mint<MOD> y = 1;
for (uint64_t i = 1; i and (i <= k); i <<= 1) {
if (k & i) y *= x;
x *= x;
}
return y;
}
};

constexpr int MOD = 998244353;
int DigitRotation::sumRotations(string x) {
// prepare
int n = x.length();
vector<mint<MOD> > e(n);
mint<MOD> x1 = 0;
REP (i, n) {
e[i] = mint<MOD>(10).pow(n - i - 1);
x1 += mint<MOD>(x[i] - '0') * e[i];
}

// sum
mint<MOD> sum = 0;
auto f = [&](char c, int i) { return mint<MOD>(c - '0') * e[i]; };
REP (c, n) REP (b, c) REP (a, b) {
if (a == 0 and x[c] == '0') continue;
sum += x1
- f(x[a], a) - f(x[b], b) - f(x[c], c)
+ f(x[c], a) + f(x[a], b) + f(x[b], c);
}
return sum.data;
}