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hdu 5898 odd-even number 数位DP
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 }
hdu 5898 odd-even number 数位DP
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