SPOJ BALNUM (数位DP)

题意：求区间内出现过的奇数是偶数，出现过的偶数是奇数的个数。

析：这个题是要三进制进行操作的。dp[i][j] 表示前 i 位，状态是 j，可以用三进制来表示 0表示没有出现，1表示奇数，2表示偶数。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[22][60000];
int a[22], b[12], f[12];

void g(int val){
    for(int i = 9; i >= 0; --i){
        b[i] = val / f[i];
        val %= f[i];
    }
}

bool judge(int val){
    g(val);
    for(int i = 0; i < 10; ++i)
        if((i&1) && b[i] == 1)  return false;
        else if(!(i&1) && b[i] == 2)  return false;
    return true;
}

int cal(int num, int val){
    g(val);
    b[num] = (b[num]&1) ? 2 : 1;
    val = 0;
    for(int i = 0; i < 10; ++i)  val += b[i] * f[i];
    return val;
}

LL dfs(int pos, int val, bool is, bool ok){
    if(!pos)  return judge(val);
    LL &ans = dp[pos][val];
    if(!ok && ans >= 0)  return ans;

    LL res = 0;
    int n = ok ? a[pos] : 9;
    for(int i = 0; i <= n; ++i)
        if(is && !i)  res += dfs(pos-1, val, is, ok && i == n);
        else res += dfs(pos-1, cal(i, val), false, ok && i == n);
    return ok ? res : ans = res;
}

LL solve(LL n){
    int len = 0;
    while(n){
        a[++len] = n % 10;
        n /= 10;
    }
    return dfs(len, 0, true, true);
}

int main(){
    f[0] = 1;
    for(int i = 1; i < 10; ++i) f[i] = f[i-1] * 3;
    memset(dp, -1, sizeof dp);
    int T;  cin >> T;
    while(T--){
        LL m, n;
        cin >> m >> n;
        cout << solve(n) - solve(m-1) << endl;
    }
    return 0;
}