SPOJ LUCIFER (Digital DP)

Lucifer-lucifer numberNo Tags

Lucifer is the only human whi have defeated Ra-one in a computer game.

Ra-one are after Lucifer for revenge and G-one are there to protect him ...

All thi G-one and ra-one nonsense have disturbed lucifers life.

He wants to get Rid of Ra-one and kill him. He found that Ra-one can is killed only by the throwing Lucifer number of weapons at him.

Lucifer number shares the some properties of Ra-one Numbers Numbers and G-one Numbers

Any number was LUCIFER number if the difference between Sum of digits at even location and Sum of digits at odd location I S prime number. For eg ... for 20314210 is Lucifer number

Digits at odd location 0,2,1,0

Digits at even location 1,4,3,2

diff = (1+4+3+2)-(0+2+1+0) =10-3 = 7 ..... a prime number.

Lucifer have access to a Warehouse which have lots of weapons.

He wants to know on how many ways can he kill him.

Can you help him?

Input

First line would have a number ' t ' denoting the number of test cases.

Each of the following T lines would have 2 numbers ' a ', ' B '

Output

Print single number per test case, depicting the count of Lucifer numbers in the range A, B inclusive.

Example

Input:
5
200 250
150 200
100 150
50 100
0 50

Output:
2
16
3
18
6


t be is less than 100

/* Link: http://www.spoj.com/problems/LUCIFER/en/Test instructions: To find the number of numbers in an interval minus the odd position in a range of prime numbers: Digital DP Dp[i][j][k] The i-bit, the odd digit and the J, Even digit and k conditions */#include <iostream> #include <cstdio> #include <cstring> #include <algorithm># include<cmath> #include <queue> #include <stack> #include <vector> #include <set># Include<map> #define L (x) (x<<1) #define R (x) (x<<1|1), #define MID (x, y) ((x+y) >>1) #define Bug printf ("hihi\n") #define EPS 1e-8typedef Long long ll;using namespace std; #define N 11int dp[n][90][90];int bit[n];int pri[    200];void Inint () {int i,j;    Pri[1]=1;    Pri[0]=1; for (i=2;i<200;i++) if (!pri[i]) for (j=i*2;j<200;j+=i) pri[j]=1;}        int dfs (int pos,int odd,int Even,bool bound) {if (pos==0) {if (even<=odd) return 0;        if (pri[even-odd]) return 0;    return 1;    } if (!bound&&dp[pos][odd][even]>=0) return Dp[pos][odd][even]; int Up=bound?    Bit[pos]:9;    int ans=0; for (int i=0;i<=up;i++) {int TT;        if (pos&1) Tt=dfs (pos-1,odd+i,even,bound&&i==up);        else Tt=dfs (pos-1,odd,even+i,bound&&i==up);    Ans+=tt;   } if (!bound) Dp[pos][odd][even]=ans; return ans;}    int solve (int x) {int i,j,len=0;        while (x) {bit[++len]=x%10;    x/=10; } return Dfs (len,0,0,true);}    int main () {int i,j,t;    Memset (Dp,-1,sizeof (DP));    Inint ();    scanf ("%d", &t);    int x, y;        while (t--) {scanf ("%d%d", &x,&y);    printf ("%d\n", Solve (y)-solve (x-1)); } return 0;}

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SPOJ LUCIFER (Digital DP)