K - Balanced Numbers

题目链接：https://vjudge.net/contest/365059#problem/K

 

题目大意：

每个奇数出现的次数是偶数，每个偶数出现的次数是奇数的数的个数

 

想法：

其实这题可以直接对每个位数都开一个大小为 2 的数组直接水过去。但是这并不是正解

正解是我们同样需要进行状态压缩把这个数转化为三进制数，0 代表未出现，1 代表出现奇数，2 代表出现偶数次

 

#pragma GCC optimize(3,"Ofast","inline")//O3优化
#pragma GCC optimize(2)//O2优化
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

LL L,R;
int b[25];
int len;
LL mem[25][60000];


bool is_ok(int x) {
    int num[10];
    for (int i = 0;i <= 9;i++) {
        num[i] = x % 3;
        if (num[i] == 2 && i % 2 == 0)
            return 0;
        if (num[i] == 1 && i % 2 == 1)
            return 0;
        x /= 3;
    }
    return 1;
}
int update(int x,int n) {
    int num[10];
    for (int i = 0;i <= 9;i++) {
        num[i] = x % 3;
        x /= 3;
    }
    if (num[n] == 0)
        num[n] = 1;
    else
        num[n] = 3-num[n];
    int ans = 0,xx = 1;
    for (int i = 0;i <= 9;i++) {
        ans += num[i] * xx;
        xx *= 3;
    }
    return ans;
}

LL dfs(int cur,int k,bool f,bool g) {
    if (cur < 0) {
        if (is_ok(k))
            return 1;
        else
            return 0;
    }
    if (!f && mem[cur][k] != -1)
        return mem[cur][k];
    int v = 9;
    if (f)
        v = b[cur];
    LL ans = 0;
    for (int i = 0;i <= v;i++) {
        if (g) {
            if (i == 0)
                ans += dfs(cur-1,0,f&&(i==v),1);
            else
                ans += dfs(cur-1,update(k,i),f&&(i==v),0);
        }
        else
            ans += dfs(cur-1,update(k,i),f&&(i==v),0);
    }
    if (!f)
        mem[cur][k] = ans;
    return ans;
}

LL solve(LL x) {
    len = 0;
    while (x) {
        b[len++] = x % 10;
        x /= 10;
    }
    return dfs(len-1,0,1,1);
}

int main() {
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    memset(mem,-1, sizeof(mem));
    while (T--) {
        cin >> L >> R;
        cout << solve(R) - solve(L-1) << endl;
    }
    return 0;
}