[DP] SRM 452 Div1 Hard IncreasingNumber

SolutionSolution

刚开始自己脑补了一个O((123)m2logn)O((123)m2log⁡n)的做法。。大概就是dpi,lo,hi,jdpi,lo,hi,j表示考虑ii位，以lolo开头以hihi结尾，模mm为jj的方案数。然后倍增DP。然而过不去。。。。。。。。。。。。。。。

可以发现对于一个数i,1≤i<10i,1≤i<10只要选10ai−1910ai−19这样的东西就好了。

这个1111....1111111....111模mm可能是有循环节的？？我是倍增DP出剩余系每个数的个数。

然后背包就好了。

要注意的是最长的那个1111.1111111.111是一定要选的。

爆int+1

// BEGIN CUT HERE

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#line 5 "IncreasingNumber.cpp"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 555;
const int MOD = 1000000007;
typedef pair<int, int> pairs;

ll n;
int m, ans, tmp;
int dp

[10];
int f
, f1
;
int val
[10];
int inv
;

class IncreasingNumber {
public:

inline void pre(int n) {
inv[1] = 1;
for (int i = 2; i <= n; i++)
inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;
}
inline int C(ll n, int m) {
int res = 1; n %= MOD;
for (int i = 0; i < m; i++) {
res = (n - i) * res % MOD;
res = (ll)res * inv[i + 1] % MOD;
}
return (res + MOD) % MOD;
}
inline void add(int &x, int a) {
x += a; while (x >= MOD) x -= MOD;
}
inline pairs solve(ll n) {
if (n == 0) return pairs(0, 0);
if (n == 1) {
f[1 % m] = 1;
return pairs(1 % m, 1 % m);
}
if (n & 1) {
pairs p = solve(n - 1);
int base = p.first * 10 % m, _base = (p.second * 10 + 1) % m;
add(f[_base], 1);
return pairs(base, _base);
} else {
pairs p = solve(n / 2);
int base = p.first * 10, _base = p.second;
for (int i = 0; i < m; i++) f1[i] = f[i];
for (int i = 0; i < m; i++)
add(f[(i * base + _base) % m], f1[i]);
return pairs(base * base / 10 % m, _base * (base + 1) % m);
}
}

int countNumbers(long long digits, int divisor) {
n = digits; m = divisor; pre(10);
tmp = (solve(n - 1).second * 10 + 1) % m;
for (int i = 0; i < m; i++)
for (int j = 0; j < 10; j++)
val[i][j] = C(f[i] + j - 1, j);
dp[0][0][0] = 1;
for (int i = 0; i < m; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < 10; k++)
for (int x = 0; x + k < 10; x++)
add(dp[i + 1][(j + i * x) % m][k + x], (ll)val[i][x] * dp[i][j][k] % MOD);
for (int i = 1; i < 10; i++)
for (int j = 0; j + i < 10; j++)
add(ans, dp[m][(m - tmp * i % m) % m][j]);
return ans;
}

// BEGIN CUT HERE
public:
void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
private:
template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
void test_case_0() { long long Arg0 = 2LL; int Arg1 = 12; int Arg2 = 4; verify_case(0, Arg2, countNumbers(Arg0, Arg1)); }
void test_case_1() { long long Arg0 = 3LL; int Arg1 = 111; int Arg2 = 9; verify_case(1, Arg2, countNumbers(Arg0, Arg1)); }
void test_case_2() { long long Arg0 = 452LL; int Arg1 = 10; int Arg2 = 0; verify_case(2, Arg2, countNumbers(Arg0, Arg1)); }
void test_case_3() { long long Arg0 = 6LL; int Arg1 = 58; int Arg2 = 38; verify_case(3, Arg2, countNumbers(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main(void) {
IncreasingNumber ___test;
___test.run_test(-1);
system("pause");
}
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