odd-even number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
分析：数位dp；
dp[i][j][k]三维分别代表位置，当前数奇偶性，个数奇偶性；
注意前导0的特判；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #define rep(i,m,n) for(i=m;i<=n;i++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") const int maxn=1e5+10; const int N=5e4+10; const int M=N*10*10; using namespace std; inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline void umax(ll &p,ll q){if(p<q)p=q;} inline void umin(ll &p,ll q){if(p>q)p=q;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,k,t,num[20],pos,cas; ll dp[20][2][2],p,q; ll dfs(int pos,int x,int y,int z,int k) { if(pos<0)return k&&(x^y)==1; if(z&&k&&dp[pos][x][y]!=-1)return dp[pos][x][y]; int now=z?9:num[pos],i; ll ret=0; rep(i,0,now) { if(k&&(x^(i&1))==1&&(x^y)==0)continue; if(!i&&!k)ret+=dfs(pos-1,x,y,z||i<num[pos],k||i); else if((i&1)==x)ret+=dfs(pos-1,x,y^1,z||i<num[pos],k||i); else ret+=dfs(pos-1,x^1,1,z||i<num[pos],k||i); } return z&&k?dp[pos][x][y]=ret:ret; } ll gao(ll x) { pos=0; while(x)num[pos++]=x%10,x/=10; return dfs(pos-1,1,0,0,0); } int main() { int i,j; memset(dp,-1,sizeof(dp)); scanf("%d",&t); while(t--) { scanf("%lld%lld",&p,&q); printf("Case #%d: %lld ",++cas,gao(q)-gao(p-1)); } return 0; }