結果

問題 No.2302 Carry X Times
ユーザー Jeroen Op de Beek
提出日時 2023-05-12 22:01:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 16 ms / 2,000 ms
コード長 3,583 bytes
コンパイル時間 2,102 ms
コンパイル使用メモリ 198,452 KB
最終ジャッジ日時 2025-02-12 22:48:16
ジャッジサーバーID
(参考情報)
judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 24
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include "bits/stdc++.h"
using namespace std;
#define all(x) begin(x),end(x)
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type>
    ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#define debug(a) cerr << "(" << #a << ": " << a << ")\n";
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int mxN = 1e5+1, oo = 1e9;
const long long MD =998244353 ;
template<long long MOD=MD> struct mdint {
int d=0;
mdint () {d=0;}
mdint (long long _d) : d(_d%MOD){
if(d<0) d+=MOD;
};
friend mdint& operator+=(mdint& a, const mdint& o) {
a.d+=o.d; if(a.d>=MOD) a.d-=MOD;
return a;
}
friend mdint& operator-=(mdint& a, const mdint& o) {
a.d-=o.d; if(a.d<0) a.d+=MOD;
return a;
}
friend mdint& operator*=(mdint& a, const mdint& o) {
return a = mdint((ll)a.d*o.d);
}
mdint operator*(const mdint& o) const {
mdint res = *this;
res*=o;
return res;
}
mdint operator+(const mdint& o) const {
mdint res = *this;
res+=o;
return res;
}
mdint operator-(const mdint& o) const {
mdint res = *this;
res-=o;
return res;
}
mdint operator^(long long b) const {
mdint tmp = 1;
mdint power = *this;
while(b) {
if(b&1) {
tmp = tmp*power;
}
power = power*power;
b/=2;
}
return tmp;
}
friend mdint operator/=(mdint& a, const mdint& o) {
a *= (o^(MOD-2));
return a;
}
mdint operator/(const mdint& o) {
mdint res = *this;
res/=o;
return res;
}
bool operator==(const mdint& o) { return d==o.d;}
bool operator!=(const mdint& o) { return d!=o.d;}
friend istream& operator>>(istream& c, mdint& a) {return c >> a.d;}
friend ostream& operator<<(ostream& c, const mdint& a) {return c << a.d;}
};
using mint = mdint<MD>;
vi digs;
int x;
mint dp[2][2][20][20][2];
bool vis[2][2][20][20][2];
mint solve(bool ba, bool bb, int i, int carry, bool curc) {
if(carry<0) return 0;
if(i==-1) {
if(curc) return 0;
if(carry==0) return 1;
return 0;
}
mint& ans = dp[ba][bb][i][carry][curc];
if(vis[ba][bb][i][carry][curc]) return ans;
vis[ba][bb][i][carry][curc]=1;
ans=0;
int la = ba?digs[i]:9, lb = bb?digs[i]:9;
for(int da=0;da<=la;++da) {
for(int db=0;db<=lb;++db) {
// assume carried:
auto res = solve(ba and la==da,bb and lb==db,i-1,carry-curc,1);
if(curc== (da+db+1>=10)) ans+=res;
res = solve(ba and la==da,bb and lb==db,i-1,carry-curc,0);
if(curc == (da+db>=10)) ans+=res;
}
}
return ans;
}
void testcase() {
string n; cin >> n;
reverse(all(n));
digs.resize(n.size());
for(int i=0;i<n.size();++i) {
digs[i]=n[i]-'0';
}
for(int i=0;i<2;++i) for(int j=0;j<2;++j) for(int k=0;k<20;++k) for(int o=0;o<20;++o) for(int l=0;l<2;++l) {
vis[i][j][k][o][l]=0;
}
cin >> x;
cout << solve(1,1,n.size()-1,x,1)+solve(1,1,n.size()-1,x,0) << '\n';
}
int main() {
int t; cin >> t;
while(t--) testcase();
}
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