数位DP
题解：http://www.cnblogs.com/algorithms/archive/2012/09/02/2667637.html
dfs的地方没太看懂……（也就那里是重点吧喂！）挖个坑……回头再看看
1 //HDOJ 3709 2 #include<cmath> 3 #include<vector> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 #include<iostream> 8 #include<algorithm> 9 #define rep(i,n) for(int i=0;i<n;++i) 10 #define F(i,j,n) for(int i=j;i<=n;++i) 11 #define D(i,j,n) for(int i=j;i>=n;--i) 12 #define pb push_back 13 using namespace std; 14 int getint(){ 15 int v=0,sign=1; char ch=getchar(); 16 while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();} 17 while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();} 18 return v*=sign; 19 } 20 const int N=20,M=1400; 21 typedef long long LL; 22 /******************tamplate*********************/ 23 24 LL dp[N][N][M]; 25 int bit[N]; 26 LL dfs(int pos,int o,int sum,int f){ 27 if (sum<0) return 0; 28 if (pos==0) return sum==0; 29 if (!f && dp[pos][o][sum]!=-1) return dp[pos][o][sum]; 30 31 LL ret=0; 32 int max=f ? bit[pos] : 9; 33 F(i,0,max) 34 ret+=dfs(pos-1,o,sum+(pos-o)*i,f && i==max); 35 if (!f) dp[pos][o][sum]=ret; 36 return ret; 37 } 38 39 LL cal(LL n){ 40 int len=0; 41 for(;n;n/=10) bit[++len]=n%10; 42 LL ret=0; 43 F(o,1,len) 44 ret+=dfs(len,o,0,1); 45 return ret-len+1; 46 } 47 int main(){ 48 #ifndef ONLINE_JUDGE 49 freopen("3709.in","r",stdin); 50 freopen("3709.out","w",stdout); 51 #endif 52 memset(dp,-1,sizeof dp); 53 int T=getint(); 54 while(T--){ 55 LL x,y; 56 cin >> x >> y; 57 cout << cal(y)-cal(x-1)<<endl; 58 } 59 return 0; 60 }