hdu 5898 odd-even number （数位dp 沈阳网络赛）

#include<bits/stdc++.h>
using namespace std;
const int N=30;
typedef unsigned long long ll;
ll f
[2]
;
// 数位、奇偶、长度
int bit
;
//位置，前导是0，奇偶，奇偶的长度，当前序列是否符合条件，最大范围
ll dfs(int pos,int pre,int x,int length,int ok,bool lim) {
if(pre&&!ok)return 0;
if(pos==0)return (length+x)%2;

if(!lim&&f[pos][x][length]!=-1)return f[pos][x][length];
int num=lim?bit[pos]:9;
ll ans=0;
for(int i=0; i<=num; i++) {
if(pre==0&&i==0)ans+=dfs(pos-1,0,0,0,1,0);
else {
if(pre==0)
ans+=dfs(pos-1,1,i%2,1,1,lim&&(i==num));
else {
if(i%2==x)ans+=dfs(pos-1,1,x,length+1,ok ,lim&&(i==num));
else ans+=dfs(pos-1,1,!x,1,ok&&(length+x)%2,lim&&(i==num));
}
}
}
if(!lim)f[pos][x][length]=ans;
return ans;
}
ll solve(ll a) {
int pos=0;
while(a) {
bit[++pos]=a%10;
a/=10;
}
return dfs(pos,0,0,0,0,1);
}
int main() {
//freopen("f.txt","r",stdin);
int T;
scanf("%d",&T);
ll a,b;
int cas=0;
memset(f,-1,sizeof(f));
while(T--) {
scanf("%lld%lld",&a,&b);
printf("Case #%d: %lld\n",++cas,solve(b)-solve(a-1));
}
return 0;
}
/*
Sample Input
2
1 100
110 220

Sample Output
Case #1: 29
Case #2: 36
*/