結果
問題 | No.315 世界のなんとか3.5 |
ユーザー |
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提出日時 | 2015-12-08 00:23:55 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 258 ms / 2,000 ms |
コード長 | 3,225 bytes |
コンパイル時間 | 985 ms |
コンパイル使用メモリ | 86,436 KB |
実行使用メモリ | 29,568 KB |
最終ジャッジ日時 | 2024-09-14 18:39:54 |
合計ジャッジ時間 | 5,621 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
other | AC * 36 |
ソースコード
#include <string>#include <vector>#include <algorithm>#include <numeric>#include <set>#include <map>#include <queue>#include <iostream>#include <sstream>#include <cstdio>#include <cmath>#include <ctime>#include <cstring>#include <cctype>#include <cassert>#include <limits>#include <functional>#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))#if defined(_MSC_VER) || __cplusplus > 199711L#define aut(r,v) auto r = (v)#else#define aut(r,v) __typeof(v) r = (v)#endif#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)#define all(o) (o).begin(), (o).end()#define pb(x) push_back(x)#define mp(x,y) make_pair((x),(y))#define mset(m,v) memset(m,v,sizeof(m))#define INF 0x3f3f3f3f#define INFL 0x3f3f3f3f3f3f3f3fLLusing namespace std;typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }template<int MOD>struct ModInt {static const int Mod = MOD;unsigned x;ModInt() : x(0) {}ModInt(signed sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }ModInt(signed long long sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }int get() const { return (int)x; }ModInt &operator+=(ModInt that) { if((x += that.x) >= MOD) x -= MOD; return *this; }ModInt &operator-=(ModInt that) { if((x += MOD - that.x) >= MOD) x -= MOD; return *this; }ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }ModInt operator+(ModInt that) const { return ModInt(*this) += that; }ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }};typedef ModInt<1000000007> mint;vector<int> digits;int P;vector<mint> memo;bool eqOK;mint rec(int i, int m, bool three, bool lt) {mint &r = memo[((i + 1) * 3 + m) * 4 + three * 2 + lt];if(r.x != -1) return r;if(i <= 4) {r = mint();int p = 1;int es = 0;rep(j, i + 1) {es += p * digits[j];p *= 10;}rep(ds, p) if(lt || ds <= es) {bool nlt = lt || ds < es;int nm = (m * p + ds) % 3;bool nthree = three;int t = ds;rep(j, i+1) {nthree |= t % 10 == 3;t /= 10;}bool ok = true;ok &= nlt || eqOK;ok &= nm == 0 || nthree;ok &= ds % P != 0;if(ok) r += 1;}return r;}r = mint();int e = digits[i];rep(d, 10) if(lt || d <= e)r += rec(i-1, (m * 10 + d) % 3, three || d == 3, lt || d < e);return r;}mint solve(const char *A) {int len = strlen(A);digits.resize(len);rep(i, len)digits[i] = A[len - 1 - i] - '0';mint r;mint undef; undef.x = -1;memo.assign((len+1) * 3 * 4, undef);return rec(len - 1, 0, false, false);}int main() {char *A = new char[200002], *B = new char[200002];while(~scanf("%s%s%d", A, B, &P)) {mint ans;eqOK = true;ans += solve(B);eqOK = false;ans -= solve(A);printf("%d\n", ans.get());}return 0;}