CodeForces 507D The Maths Lecture constructs EXACTLY n digits and there is a suffix that can be divisible by k dp, codeforces507d

CodeForces 507D The Maths Lecture constructs EXACTLY n digits and there is a suffix that can be divisible by k dp, codeforces507d

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D. The Maths Lecturetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. but one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn' t.

First he gave Amr two positive integersNAndK. Then he asked Amr, how many integer numbersXLatency> limit 0 exist such that:

• Decimal representationX(Without leading zeroes) consists of exactlyNDigits;
• There exists some integerYLatency> limit 0 such that:
  • ;
  • Decimal representationYIs a suffix of decimal representationX.

As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a numberM.

Can you help Amr escape this embarrassing situation?

Input

Input consists of three integersN, Bytes,K, Bytes,M(1 digit ≤ DigitNLimit ≤ limit 1000, 1 limit ≤ limitKLimit ≤ limit 100, 1 limit ≤ limitMLimit ≤ limit 109 ).

Output

Print the required number moduloM.

Sample test (s) input

1 2 1000

Output

4

Input

2 2 1000

Output

45

Input

5 3 1103

Output

590

Note

A suffix of a stringSIs a non-empty string that can be obtained by removing some number (possibly, zero) of first characters fromS.

Question:

Given n, k, m

The number of integers EXACTLY n digits. A suffix of this integer can be divisible by k. The answer is m.

Ideas:

Because it is a suffix, it is constructed from low to high.

Dp [I] [j] [0] indicates an integer with an I-bit length. The result of mod k is j, and there is no number of suffixes that can divide k.

Dp [I] [j] [1] indicates an integer with an I-bit length. The result of mod k is j, and there is a number of suffixes that can divide k.

Then, add a number before a certain State to determine whether the number can be divisible by k.

#include <cstdio>#include <algorithm>#include <string.h>#include <queue>#include <cstring>#include <cmath>#include <iostream>#include <vector>using namespace std;#define y1 Y1typedef long long ll;const int N = 1005;ll ten[N], dp[N][100][2];ll n, k, m;void add(ll &x, ll y){x += y;if (x >= m)x -= m;}int main(){while (cin >> n >> k >> m){ten[0] = 1 % m;for (int i = 1; i < N; i++)ten[i] = (ten[i - 1] * 10LL) % k;memset(dp, 0, sizeof dp);dp[0][0][0] = 1;for (int i = 0; i < n; i++){for (int j = 0; j < k; j++){for (int z = 0; z < 10; z++){if (i == n - 1 && z == 0)continue;ll now = (z*ten[i] + j) % k;if (now == 0 && z){add(dp[i + 1][now][1], dp[i][j][0]);}elseadd(dp[i + 1][now][0], dp[i][j][0]);add(dp[i + 1][now][1], dp[i][j][1]);}}}ll ans = 0;for (int i = 0; i < k; i++)ans += dp[n][i][1];cout << ans%m << endl;}return 0;}