[Hihocoder 1033] staggered and digital DP/Memory search

#1033: crossover and time limit: Memory ms single point time limit: 256 Ms memory limit: MB

Description

Given a number X, it is set to A0 in sequence from the decimal position to the lower position ,? A1 ,?...,? An? -? 1. Define the interlaced and function:

F (x )? =? A0? -? A1? +? A2? -?...? +? (? -? 1) n? -? 1An? -? 1

For example:

F (3214567 )? =? 3? -? 2? +? 1? -? 4? +? 5? -? 6? +? 7? =? 4

Given

Input

Only one row of input data contains three integers: l ,? R ,? K (0? ≤? L? ≤? R? ≤? 1018 ,? | K |? ≤? 100 ).

Output

Output a row of an integer to indicate the result. Considering that the answer may be large, the output result is modeled as 109? +? 7.

Prompt

For example, if the number of conditions is 110 and 121, the result is 231 = 110 + 121.

More examples:

Input

4344 3214567 3

Output

611668829

Input

404491953 1587197241 1

Output

323937411

Input

60296763086567224 193422344885593844 10

Output

608746132

Input

100 121-1

Output

120

Sample Input

100 121 0

Sample output

231

Question = _ = the question comes from the first challenge of hihocoder.

At the beginning, I thought it was a number theory topic, and later I found it was a number of DP, which is easy to hold:

1. During the memory-based search and writing, the same number and number must be staggered, and the numbers and numbers of the same staggered number must be respectively DP

2. The calculation method of one digit and two digits is different. We need to discuss the calculation method of the two digits.

3. Because the result may be relatively large, we need to use the same remainder theorem for each step to prevent the long from popping up during the operation.

The concept of memory-based search,

The sum of the current staggered and the same number (number in the to-be-searched status and + the size of the number currently searched * The number of currently searched qualified numbers ).

# Include <cstdio> # include <cstring> long mod = 1000000007; long base [20]; long l, R, K, bit [20], BT, YY; struct node {long S, N; // s represents the number and N represents the number}; node DP [20] [400]; // status transition node DFS (long POs, long target, long limit) // digital DP, which is basically a template {node T; T. S = T. n = 0; If (Pos = 0) {// process to the last bit and directly judge that the returned if (target = 100) T. n = 1; return t;} If (Limit = 0) & (DP [POS] [target]. n! =-1) return DP [POS] [target]; long tail = limit? Bit [POS]: 9; long SGN = (yy-Pos) % 2 )? (-1) :( 1); // determine the long head symbol; If (Pos = YY) Head = 1; else head = 0; // determine the start and end of the search for (INT I = head; I <= tail; I ++) {node TMP = DFS (pos-1, target-I * SGN, (Limit = 1) & (I = bit [POS]); If (TMP. n)> 0) {T. N + = TMP. n; Long Q; q = (TMP. N % mod) * base [POS]) % mod) * I) % MOD; // result Cool T. S + = (TMP. s) % MOD; T. S % = MOD; T. S + = Q; T. S % = MOD; // each step must be the same.} If (Limit = 0) DP [POS] [target] = T; return t ;} long long CAL (long X, long y) {long ans = 0; If (x =-1) return 0; If (x = 0) return 0; bt = 0; while (x) {BT ++; bit [BT] = x % 10; X/= 10;} For (yy = 1; YY <= Bt; YY ++) {memset (DP,-1, sizeof DP); ans + = DFS (YY, Y + 100, YY = BT ). s; // process the number with a length of YY ans = (ANS + mod) % MOD;} return ans;} int main () {base [1] = 1; for (INT I = 2; I <= 19; I ++) base [I] = (base [I-1] * 10) % MOD; scanf ("% LLD", & L, & R, & K); {printf ("% LLD", (CAL (R, K) -Cal (L-1, k) + mod) % mod);} return 0 ;}