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;}