hihoCoder 1259 A Math Problem（数位dp）

题目链接：hihoCoder 1259 A Math Problem

解题思路

公式推导完为f(2n) = 3 * f(n)， f(2n+1) = 3 * f(n) + 1，即为i的2进制作为3进制计算贡献。

代码

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 65540;
const int maxm = 65;

int K;
ll N, dp[maxm][maxn];

ll solve () {
    int s = 0, bit[maxm];
    for (int i = 60; i >= 0; i--) {
        bit[i] = N % 2;
        N /= 2;
    }

    memset(dp, 0, sizeof(dp));
    for (int i = 0; i <= 60; i++) {
        for (int j = 0; j < bit[i]; j++)
            dp[i][(s*3+j)%K]++;
        s = (s * 3 + bit[i]) % K;

        for (int j = 0; j < K; j++) {
            dp[i+1][j*3 % K] += dp[i][j];
            dp[i+1][(j*3+1)%K] += dp[i][j];
        }
    }
    dp[60][s]++;
    dp[60][0]--;

    ll ans = 0;
    for (int i = 0; i < K; i++)
        ans ^= dp[60][i];
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%lld%d", &N, &K);
        printf("%lld\n", solve());
    }
    return 0;
}