HDU 5898 odd-even number(数位dp)

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;
#define   MAX           40005
#define   MAXN          6005
#define   maxnode       15
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
//const double inf   = 1e18;
const double eps   = 1e-8;
const LL     mod   = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

int a[25];
LL dp[25][25][25];

LL dfs(int len,int numo,int nume,int fp,int z){
if(len==0){
if(z) return 0;
if(numo%2==0&&nume==0) return 1;
if(numo==0&&nume%2) return 1;
return 0;
}
if(!fp&&!z&&dp[len][numo][nume]!=-1) return dp[len][numo][nume];
LL ans=0;
int cnt=fp?a[len]:9;
for(int i=0;i<=cnt;i++){
if(i==0&&z) ans+=dfs(len-1,numo,nume,0,1);
else{
if(i%2&&((nume%2)||nume==0)) ans+=dfs(len-1,numo+1,0,fp&&(i==cnt),0);
else if(i%2==0&&((numo%2==0)||(numo==0))) ans+=dfs(len-1,0,nume+1,fp&&(i==cnt),0);
}
}
if(!fp&&!z) dp[len][numo][nume]=ans;
return ans;
}

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

int main(){
int t,kase=0;
cin>>t;
while(t--){
LL L,R;
scanf("%I64d%I64d",&L,&R);
kase++;
mem(dp,-1);
//cout<<solve(R)<<endl;
//cout<<solve(L-1)<<endl;
printf("Case #%d: %I64d\n",kase,solve(R)-solve(L-1));
}
return 0;
}