Spoj NUMTSN NUMTSN-369 Numbers (digital DP)

NUMTSN-369 NumbersNo Tags

7.369 numbers

A number is said to be a 369 number if

1. The count of 3s is equal to count of 6s and the count of 6s are equal to count of 9s.
2. The count of 3s is at least 1.

For Example 12369, 383676989, 396 all is 369 numbers whereas 213, 342143, 111 is not.

Given A and B find how many 369 numbers is there in the interval [a, b]. Print the answer modulo 1000000007.

Input

The first line contains the number of test cases (T) followed to T lines each containing 2 integers A and B.

Output

For each test case output the number of 369 numbers between A and B inclusive.

Constraints

t<=100

1<=a<=b<=10^50

Sample Input

3

121 4325

432 4356

4234 4325667

Sample Output

60

58

207159

Links: http://www.spoj.com/problems/NUMTSN/en/

/ *
Intent: find the number of the same number (and at least 1) in a range of 3 6 9
Idea: Note that the number is very large, with a power of 10 ^ 50, so save it with a string
     The digits dp dp [i] [j] [k] [m] represent the numbers up to the i-th position, 3 6 9


* /

#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-8

typedef long long ll;

using namespace std;
#define N 55
#define mod 1000000007

ll dp [N] [N] [N] [N]; //
int bit [N];

void change (char * a)
{
    int i, j;
    int len = strlen (a);
    len--;
    while (len> = 0)
    {
        a [len]-;
        if (a [len]> = '0') break;
        a [len] = '9';
        len--;
    }
    len = strlen (a);
    for (i = 0; i <len-1; i ++) if (a [i]> = '1') break;
    int k = 0;
    for (i; i <len; i ++)
        a [k ++] = a [i];
    a [k] = '\ 0';
}

ll dfs (int pos, int le, int mid, int ri, bool bound)
{
    if (pos == 0) return le == mid && mid == ri && le> = 1? 1: 0;
    if (! bound && dp [pos] [le] [mid] [ri]> = 0) return dp [pos] [le] [mid] [ri];
    int up = bound? bit [pos]: 9;
    ll ans = 0;
    for (int i = 0; i <= up; i ++)
    {
        if (i == 3)
          ans = (ans + dfs (pos-1, le + 1, mid, ri, bound && i == up))% mod;
        else
          if (i == 6)
           ans = (ans + dfs (pos-1, le, mid + 1, ri, bound && i == up))% mod;
        else
            if (i == 9)
             ans = (ans + dfs (pos-1, le, mid, ri + 1, bound && i == up))% mod;
        else
             ans = (ans + dfs (pos-1, le, mid, ri, bound && i == up))% mod;
    }
    if (! bound) dp [pos] [le] [mid] [ri] = ans;
    return ans;
}


ll solve (char * c)
{
   int i, j;
   int len = strlen (c);
   for (i = 1; i <= len; i ++)
    bit [i] = c [len-i]-'0';

   return dfs (len, 0,0,0, true)% mod;
}

int main ()
{
    int i, j, t;
    char a [N], b [N];
    memset (dp, -1, sizeof (dp));
    scanf ("% d", & t);
    while (t--)
    {
        scanf ("% s% s", a, b);
        change (a);
        printf ("% lld \ n", (solve (b) -solve (a) + mod + mod)% mod); // must + mod% mod because it may be negative
    }
    return 0;
} 

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Spoj NUMTSN NUMTSN-369 Numbers (digital DP)