CF855E Salazar Slytherin‘s Locket(数位dp 状压)

传送门

分析

数位dp 问题是如何记录状态。
题目要统计出现偶数次的个数->二进制
利用二进制记录第i个数出现次数的奇偶性。

代码

#include <bits/stdc++.h>

using namespace std;
//-----pre_def----
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
#define fir(i, a, b) for (int i = (a); i <= (b); i++)
#define rif(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
#define init_h memset(h, -1, sizeof h), idx = 0;
#define lowbit(x) x &(-x)

//---------------
const int N = 1e5 + 10;
int n, m, b;
LL l, r;
int num[80], len;
LL f[11][80][(1 << 11)-1];

LL dfs(int pos, bool limit, bool lead, int st)
{
    LL ans = 0;
    if (pos == 0)
        return (st == (1 << b) - 1); //二进制位上 1为出现偶数次
    if (!limit && !lead && f[b][pos][st] != -1)
        return f[b][pos][st];
    int up = limit ? num[pos] : b - 1;
    fir(i, 0, up)
    {
        ans += dfs(pos - 1, limit && (i == up), lead && (i == 0),
                   (lead && i == 0) ? st : (1 << i) ^ st);
    }
    if (!limit && !lead)
        f[b][pos][st] = ans;
    return ans;
}

LL solve(LL x)
{
    len = 0;

    while (x)
    {
        num[++len] = x % b;
        x /= b;
    }
    return dfs(len, 1, 1, (1 << b) - 1);
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    int StartTime = clock();
#endif
    int T;
    scanf("%d", &T);
    memset(f, -1, sizeof f);
    while (T--)
    {
        scanf("%d%lld%lld", &b, &l, &r);
        printf("%lld\n", solve(r) - solve(l - 1));
    }
#ifndef ONLINE_JUDGE
    printf("Run_Time = %d ms\n", clock() - StartTime);
#endif
    return 0;
}