題目鏈接
Given an integer k and a number of bits b (1 ≤ b ≤ 128), calculate the total number of 1 bits in thebinary representations of multiples of k between 0 and $2^b$? 1 (inclusive), modulo 1,000,000,009.

Input

The input will consist of two integers k and b on a single line, with 1 ≤ k ≤ 1000 and 1≤ b ≤ 128.

Output

Write your result as an integer on a single line.

題意

求0~$2^b$中所有k的倍數二進制表示1的個數

思路

數位dp:

• dp[ i ][ j ] 表示到第i個二進制(0~$(2^1?1)$)，中模K余數為j的二進制表示1的個數
• cnt[ i ][ j ] 表示到第i個二進制(0~$(2^1?1)$)，中模K余數為j的數的個數

狀態轉移：
j = $(2^i+pre)\%K$

• dp[ i ][ j ] = dp[ i - 1 ][ j ] + dp[ i - 1 ][ pre ] + cnt[ i - 1 ][ pre ]
• cnt[ i ][ j ] = cnt[ i - 1 ][ j ] + cnt[ i - 1 ][ pre ]
  Ans = dp[ n ][ 0 ]

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i <= n; ++i) #define maxn 1005 #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 using namespace std; // __int128 read() { __int128 x = 0, f = 1; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); } while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); } return x * f;} // void print(__int128 x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) print(x / 10); putchar(x % 10 + '0');} // p1 not equal, p2 equal int inf = 0x3f3f3f3f; const LL mod = 1e9 + 9; LL dp[200][1003], cnt[200][1003];int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout); #endifios::sync_with_stdio(false);cin.tie(0); cout.tie(0);LL k, b;while (cin >> k >> b) {mem(cnt, 0);mem(dp, 0);cnt[1][0]++;cnt[1][1%k]++;dp[1][1%k]++;LL tmp = 1 % k;for (int i = 2; i <= b; ++i) {tmp = tmp * 2 % k;for (int j = 0; j < k; ++j) {int pre = (j + k - tmp) % k;cnt[i][j] = (cnt[i-1][j] + cnt[i-1][pre]) % mod;dp[i][j] = (dp[i-1][j] + cnt[i-1][pre] + dp[i-1][pre] % mod) % mod;}}cout << dp[b][0] << endl;}return 0; }

總結

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