#965459 (C++17) No.2699 Simple Math (Returned)

• 問題: No.2699 Simple Math (Returned)
• ユーザー: GOTKAKO
• 提出日時: 2024-03-29 21:13:50
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 485 ms / 2,000 ms
• コード長: 2,178 bytes
• コンパイル時間: 1,799 ms
• コンパイル使用メモリ: 193,484 KB
• 最終ジャッジ日時: 2025-02-20 14:42:58
• ジャッジサーバーID （参考情報）: judge2 / judge3

ファイルパターン	結果
sample	AC * 1
other	AC * 11

ソースコード

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#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
//入力が必ずmod未満の時に使う.
struct mint{
    long long v = 0;
    mint(){} mint(int a){v = a;} mint(long long a){v = a;} mint(unsigned long long a){v = a;}
    long long val(){return v;}
    void modu(){v %= mod;}
 
    mint repeat2mint(long long a,long long b){
        mint ret = 1,p = a;
        while(b){if(b&1) ret *= p; p *= p; b >>= 1;}
        return ret;
    };
 
    mint operator+(mint b){return (v+b.v)%mod;}
    mint operator-(mint b){return (v-b.v+mod)%mod;}
    mint operator*(mint b){return v*b.v%mod;}
    mint operator/(mint b){
        if(b.v == 0) assert(false);
        return v*(repeat2mint(b.v,mod-2)).v%mod;
    }
 
    void operator+=(mint b){v = (v+b.v)%mod;}
    void operator-=(mint b){v = (v-b.v+mod)%mod;}
    void operator*=(mint b){v = v*b.v%mod;}
    void operator/=(mint b){
    if(b.v == 0) assert(false);
        v = v*repeat2mint(b.v,mod-2).v%mod;
    }
 
    void operator++(int){v = (v+1)%mod; return;}
    void operator--(int){v = (v-1+mod)%mod; return;}
    bool operator==(mint b){if(v == b.v) return true; else return false;}
    bool operator!=(mint b){if(v != b.v) return true; else return false;}
    bool operator>(mint b){if(v > b.v) return true; else return false;}
    bool operator>=(mint b){if(v >= b.v) return true; else return false;}
    bool operator<(mint b){if(v < b.v) return true; else return false;}
    bool operator<=(mint b){if(v <= b.v) return true; else return false;}
 
    mint pow(long long x){return repeat2mint(v,x);}
    mint inv(){return mint(1)/v;}
};
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T; cin >> T;
    while(T--){
        int N,M; cin >> N >> M;
        int n = N;
        N %= 2*M;
        if(N <= M) cout << (mint(10).pow(N)-1).v << endl;
        else{
            mint now = mint(10).pow(M);
            now -= mint(10).pow(N-M);
            cout << now.v << endl;
        } 
        continue;
        long long c1 = 0,c2 = 1;
        for(int i=0; i<n; i++) c1 *= 10,c1 += 9;
        for(int i=0; i<M; i++) c2 *= 10;
        c2++; cout << c1%c2 << endl;
    }
}
 
 
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