[HDOJ5898]odd-even number（数位dp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5898

题意：求[l,r]区间内数字，满足连续奇数的个数是偶数个，连续偶数的个数是奇数个。

dp(l,pre,con,fz)表示前l位，最后一位是pre，并且此时这个pre所在的连通块已经有con个了，fz来区分是不是前导零。

#include <bits/stdc++.h>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 19;
int digit[maxn];
LL l, r;
LL dp[maxn][2][2][2];

LL dfs(int l, int pre, int con, bool fz, bool flag) {
  if(l == 0) {
    if(pre % 2 == 0) {
      if(con % 2 == 0) return 0;
      if(con % 2 == 1) return 1;
    }
    else {
      if(con % 2 == 0) return 1;
      if(con % 2 == 1) return 0;
    }
  }
  if(!flag && ~dp[l][pre][con][fz]) return dp[l][pre][con][fz];
  LL ret = 0;
  int pos = flag ? digit[l] : 9;
  if(fz) {
    Rep(i, pos+1) {
      ret += dfs(l-1, i%2, 1, fz&&(i==0), flag&&(i==pos));
    }
  }
  else if((pre + con) % 2 == 1) {
    Rep(i, pos+1) {
      if((i + pre) % 2 == 0) {
        ret += dfs(l-1, i%2, (con+1)%2, fz&&(i==0), flag&&(i==pos));
      }
      else {
        ret += dfs(l-1, i%2, 1, fz&&(i==0), flag&&(i==pos));
      }
    }
  }
  else {
    Rep(i, pos+1) {
      if((i + pre) % 2 == 0) {
        ret += dfs(l-1, i%2, (con+1)%2, fz&&(i==0), flag&&(i==pos));
      }
    }
  }
  if(!flag) dp[l][pre][con][fz] = ret;
  return ret;
}

LL f(LL x) {
  int pos = 0;
  while(x) {
    digit[++pos] = x % 10;
    x /= 10;
  }
  return dfs(pos, 0, 1, true, true);
}

signed main() {
  //FRead();
  int T, _ = 1;
  Rint(T);
  Clr(dp, -1);
  W(T) {
    cin >> l >> r;
    printf("Case #%d: ", _++);
    cout << f(r) - f(l-1) << endl;
  }
  RT 0;
}