HDU 5898 odd-even number (数位DP)

题意：给定一个区间，统计连续是奇数的个数是偶数，连续是偶数的个数是奇数的个数。

析：dp[i][j][k] 表示前 i 位，前一位是 j ，连续 k 次。

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>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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[25][2][25];
int a[20];

LL dfs(int pos, int last, int k, bool is, bool ok){
    if(!pos)  return (last + k) & 1;
    LL &ans = dp[pos][last][k];
    if(!ok && !is && ans >= 0)  return ans;

    LL res = 0;
    int n = ok ? a[pos] : 9;
    for(int i = 0; i <= n; ++i){
        if(is){
            if(!i)  res += dfs(pos-1, 0, 1, !i, ok && i == n);
            else res += dfs(pos-1, i&1, 1, false, ok && i == n);
        }
        else if(i & 1){
            if(last & 1)  res += dfs(pos-1, 1, k+1, is, ok && i == n);
            else if(k & 1)  res += dfs(pos-1, 1, 1, is, ok && i == n);
        }
        else{
            if(last % 2 == 0)  res += dfs(pos-1, 0, k+1, is, ok && i == n);
            else if(k % 2 == 0)  res += dfs(pos-1, 0, 1, is, ok && i == n);
        }
    }
    if(!ok && !is)  ans = res;
    return res;
}

LL solve(LL n){
    int len = 0;
    while(n > 0){
        a[++len] = n % 10;
        n /= 10;
    }
    return dfs(len, 0, 1, true, true);
}

int main(){
    memset(dp, -1, sizeof dp);
    LL x, y;
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%lld %lld", &x, &y);
        printf("Case #%d: %lld\n", kase, solve(y)-solve(x-1));
    }
    return 0;
}