結果

問題 No.1085 桁和の桁和
ユーザー saxofone111
提出日時 2021-03-14 22:42:37
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,614 bytes
コンパイル時間 2,094 ms
コンパイル使用メモリ 201,020 KB
最終ジャッジ日時 2025-01-19 17:10:28
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 4
other AC * 32 WA * 3
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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"
#define MOD 1000000007
#define rep(i, n) for(ll i=0; i < (n); i++)
#define rrep(i, n) for(ll i=(n)-1; i >=0; i--)
#define ALL(v) v.begin(),v.end()
#define rALL(v) v.rbegin(),v.rend()
#define FOR(i, j, k) for(ll i=j;i<k;i++)
#define debug_print(var) cerr << #var << "=" << var <<endl;
#define DUMP(i, v)for(ll i=0;i<v.size();i++)cerr<<v[i]<<" "
#define fi first
#define se second
using namespace std;
typedef long long int ll;
typedef vector<ll> llvec;
typedef vector<double> dvec;
typedef pair<ll, ll> P;
typedef long double ld;
struct edge{ll x, c;};
ll mod(ll a, ll mod){
ll res = a%mod;
if(res<0)res=res + mod;
return res;
}
ll modpow(ll a, ll n, ll mod){
ll res=1;
while(n>0){
if(n&1) res=res*a%mod;
a=a*a%mod;
n>>=1;
}
return res;
}
ll modinv(ll a, ll mod){
ll b=mod, u=1, v=0;
while(b){
ll t=a/b;
a-=t*b; swap(a, b);
u-=t*v; swap(u, v);
}
u%=mod;
if(u<0)u+=mod;
return u;
}
ll gcd(ll a, ll b){
ll r = a%b;
if(r==0) return b;
else return gcd(b, a%b);
}
// @param b 1
// @return pair(g, x) s.t. g = gcd(a, b), xa = g (mod b), 0 <= x < b/g
std::pair<long long, long long> inv_gcd(long long a, long long b) {
a = mod(a, b);
if (a == 0) return {b, 0};
// Contracts:
// [1] s - m0 * a = 0 (mod b)
// [2] t - m1 * a = 0 (mod b)
// [3] s * |m1| + t * |m0| <= b
long long s = b, t = a;
long long m0 = 0, m1 = 1;
while (t) {
long long u = s / t;
s -= t * u;
m0 -= m1 * u; // |m1 * u| <= |m1| * s <= b
// [3]:
// (s - t * u) * |m1| + t * |m0 - m1 * u|
// <= s * |m1| - t * u * |m1| + t * (|m0| + |m1| * u)
// = s * |m1| + t * |m0| <= b
auto tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
// by [3]: |m0| <= b/g
// by g != b: |m0| < b/g
if (m0 < 0) m0 += b / s;
return {s, m0};
}
// (rem, mod)
std::pair<long long, long long> crt(const std::vector<long long>& r,
const std::vector<long long>& m) {
assert(r.size() == m.size());
int n = int(r.size());
// Contracts: 0 <= r0 < m0
long long r0 = 0, m0 = 1;
for (int i = 0; i < n; i++) {
assert(1 <= m[i]);
long long r1 = mod(r[i], m[i]), m1 = m[i];
if (m0 < m1) {
std::swap(r0, r1);
std::swap(m0, m1);
}
if (m0 % m1 == 0) {
if (r0 % m1 != r1) return {0, 0};
continue;
}
// assume: m0 > m1, lcm(m0, m1) >= 2 * max(m0, m1)
// (r0, m0), (r1, m1) -> (r2, m2 = lcm(m0, m1));
// r2 % m0 = r0
// r2 % m1 = r1
// -> (r0 + x*m0) % m1 = r1
// -> x*u0*g % (u1*g) = (r1 - r0) (u0*g = m0, u1*g = m1)
// -> x = (r1 - r0) / g * inv(u0) (mod u1)
// im = inv(u0) (mod u1) (0 <= im < u1)
long long g, im;
std::tie(g, im) = inv_gcd(m0, m1);
long long u1 = (m1 / g);
// |r1 - r0| < (m0 + m1) <= lcm(m0, m1)
if ((r1 - r0) % g) return {0, 0};
// u1 * u1 <= m1 * m1 / g / g <= m0 * m1 / g = lcm(m0, m1)
long long x = (r1 - r0) / g % u1 * im % u1;
// |r0| + |m0 * x|
// < m0 + m0 * (u1 - 1)
// = m0 + m0 * m1 / g - m0
// = lcm(m0, m1)
r0 += x * m0;
m0 *= u1; // -> lcm(m0, m1)
if (r0 < 0) r0 += m0;
}
return {r0, m0};
}
bool is_prime(ll n){
ll i = 2;
if(n==1)return false;
if(n==2)return true;
bool res = true;
while(i*i <n){
if(n%i==0){
res = false;
}
i = i+1;
}
//if(i==1)res = false;
if(n%i==0)res=false;
return res;
}
#define fracMax 10000000
ll frac[fracMax];
ll ifrac[fracMax];
void gen(){
frac[0] = 1;
ifrac[0] = 1;
rep(i, fracMax-1){
frac[i+1] = mod((i+1)*frac[i], MOD);
ifrac[i+1] = modinv(frac[i+1], MOD);
}
return;
}
ll binom(ll n, ll k){
if(k<0 or k>n){
return 0;
}else{
return mod(mod(frac[n]*ifrac[n-k], MOD)*ifrac[k], MOD);
}
}
string S;
ll D;
/**************************************
** A main function starts from here **
***************************************/
int main(){
cin >> S >> D;
ll N = S.size();
if(D==0){
bool flg = true;
rep(i, N){
flg = flg and (S[i]=='0' or S[i] =='?');
}
if(flg){
cout << 1 << endl;
return 0;
}else{
cout << 0 << endl;
return 0;
}
}
vector<llvec> dp(N+1, llvec(9, 0));
dp[0][0] = 1;
rep(i, N){
rep(j, 9){
if(S[i]=='?'){
rep(k, 10){
dp[i+1][mod(j+k, 9)] = mod(dp[i+1][mod(j+k, 9)] + dp[i][j], MOD);
}
}else{
ll num = S[i] - '0';
dp[i+1][mod(j+num, 9)] = mod(dp[i+1][mod(j+num, 9)] + dp[i][j], MOD);
}
}
}
cout << dp[N][mod(D, 9)]<<endl;
return 0;
}
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