hdu 5898 odd-even number 数位DP

Posted 2020-08-11 hxer

odd-even number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 716    Accepted Submission(s): 385

Problem Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

 

 

Input

First line a t,then t cases.every line contains two integers L and R.

 

 

Output

Print the output for each case on one line in the format as shown below.

 

 

Sample Input

2 1 100 110 220

 

 

Sample Output

Case #1: 29

Case #2: 36

 

思路：数位DP的递归出口为pos = 0，这时只能判断DP状态末尾的情况，即对于构造一个数来说只能判断最低的奇偶相同的几位，那么如何才能知道中间是否有不符合的情况？

这就需要设置两个变量，一个表示第pos的奇偶性，一个表示以pos位为末尾的中间状态的长度。不过还在中间加上了一个前导0的标识zero.这个只为编码的方便，其实可以特判出是否为前导0的；

ps:说到底还是套路题....

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<queue>
  5 #include<set>
  6 #include<vector>
  7 #include<map>
  8 #include<cstring>
  9 #include<string>
 10 #include<cmath>
 11 using namespace std;
 12 #define MS1(a) memset(a, -1, sizeof(a))
 13 #define MS0(a) memset(a, 0,sizeof(a))
 14 #define rep1(i,a,b) for(int i = a; i <= b; i++)
 15 #define rep0(i,a,b) for(int i = a; i < b; i++)
 16 #define rep_0(i,a,b) for(int i = b; i > a; i--)
 17 #define rep_1(i,a,b) for(int i = b; i >= a; i--)
 18 #define inf 0x3f3f3f3f
 19 #define INF 0x3f3f3f3f3f3f3f3f
 20 #define bit(p) (1<<p)
 21 #define bitnum(a) __builtin_popcount(a)
 22 #define lowbit(x) (x&(-x))
 23 #define eps 1e-8
 24 #define mod 1000000007
 25 typedef pair<int,int> PII;
 26 #define ll long long
 27 #define ull unsigned long long
 28 #define uint unsigned int
 29 #define lson l, m, rt << 1
 30 #define rson m+1, r, rt << 1|1
 31 #define MK make_pair
 32 #define A first
 33 #define B second
 34 #define pb(a) push_back(a)
 35 #define zero(x) (((x)>0?(x):-(x))<eps)
 36 template<typename T>
 37 void read1(T &m)
 38 {
 39     T x = 0,f = 1;char ch = getchar();
 40     while(ch <‘0‘ || ch >‘9‘){ if(ch == ‘-‘) f = -1;ch=getchar(); }
 41     while(ch >= ‘0‘ && ch <= ‘9‘){ x = x*10 + ch - ‘0‘;ch = getchar(); }
 42     m = x*f;
 43 }
 44 template<class T1, class T2>
 45 void read2(T1 &a,T2 &b){read1(a);read1(b);}
 46 template<class T1, class T2, class T3>
 47 void read3(T1 &a,T2 &b,T3 &c){read1(a);read1(b);read1(c);}
 48 
 49 template<class T1, class T2>
 50 inline void gmax(T1& a, T2 b){ if(a < b) a = b; }
 51 template<class T1, class T2>
 52 inline void gmin(T1&a , T2& b){ if(a > b) a = b; }
 53 
 54 void debug(ll bug) { cout<<"**** "<<bug<<endl;}
 55 void bug(){ puts(" **** bug"); }
 56 void ok(){ puts(" **** ok"); }
 57 int bit[20];
 58 ll dp[20][2][20][2];
 59 ll dfs(int pos, int odd_even, int len, int zero, int on)
 60 {
 61     if(pos == 0) return odd_even^(len&1);
 62     ll ans = dp[pos][odd_even][len][zero];
 63     if(!on && ans != -1) return ans;
 64     int e = on?bit[pos]:9;
 65     ans = 0;
 66     for(int i = 0; i <= e; i++){
 67         if(zero == 1){
 68             if(i == 0) ans += dfs(pos-1, 0, 0, 1, 0);
 69             else ans += dfs(pos-1, i&1, 1, 0, on && i == e);
 70             //cout<<i<<" "<<ans<<endl;
 71         } else {
 72             if(i & 1){
 73                 if(odd_even & 1) ans += dfs(pos-1, odd_even, len+1, 0, on && i == e);
 74                 else if(len & 1) ans += dfs(pos-1, odd_even^1, 1, 0, on && i == e);
 75             } else {
 76                 if(odd_even % 2 == 0) ans += dfs(pos-1, odd_even, len+1, 0, on && i == e);
 77                 else if(len % 2 == 0) ans += dfs(pos-1, odd_even^1, 1, 0, on && i == e);
 78             }
 79         }
 80     }
 81     if(!on) dp[pos][odd_even][len][zero] = ans;
 82     return ans;
 83 }
 84 ll calc(ll x)
 85 {
 86     int top = 0;
 87     while(x){
 88         bit[++top] = x%10;
 89         x /= 10;
 90     }
 91     return dfs(top, 0, 0, 1, 1);
 92 }
 93 int main()
 94 {
 95     MS1(dp);
 96     int T, kase = 1;
 97     cin >> T;
 98     while(T--){
 99         ll L, R;
100         read2(L,R);
101         printf("Case #%d: %lld\n",kase++,calc(R) - calc(L-1));
102     }
103     return 0;
104 }