BOJ 316 password DP

The password question was originally known to use DP, but I don't know how to think about it. Asked the classmate, his general idea is this, assuming the string is Ch [1005], the string length is Len, set DP [0] = 1, 1-> len-1 initialization to 0. DP [I] indicates the total number of combination methods before I bit. Therefore, we require the result of DP [Len].
Example: 1 2 3
Initialize DP: 1 0 0 0
First iteration 1 (1) 1 (12) 1 (123)
Second iteration 2 (1) (2) 2 (1) (23)
Third iteration 2 (12) (3) (1) (2) (3)

The above process can be analyzed as follows:
During the first iteration, 1 <= 255, + dp [0] = 1 at the corresponding position of 2, indicating that 2 can be reached by (1); 12 <= 255, + dp [0] at the position corresponding to 3, indicating that 3 can be reached by (12); 123 <= 255, + dp [0] at the last digit, it indicates that the last digit can be reached through (123. In the second iteration, 2 <= 255, + dp [1] = 1 at the corresponding position of 3, indicating that 3 can be reached from 2; 23 <= 255, + dp [1] at the position corresponding to the last digit indicates that the last digit can be reached through 2. Similarly, we can see that the first iteration divides the data into (1) (12) (123) such a combination (both include 1), the second is divided into (2) (23), and the third is divided into (3.

For cases where 0 is included, such as: 1 0 2
Since 0 must be used as one, or can be combined with the previous one, but cannot be combined with the later one, therefore, you only need to add the corresponding value of 0 to the next bit to jump out (break ).

# Include <stdio. h> # include <stdlib. h> # include <string. h> char ch [1005]; long DP [1005]; long res; int t; int char_to_int (int from, int to) {int ans = 0; for (INT I = from; I <= to; I ++) ans = ans * 10 + (CH [I]-'0'); Return ans ;} int main () {scanf ("% d", & T); While (t --) {memset (DP, 0, sizeof (DP )); DP [0] = 1; scanf ("% s", CH); int Len = strlen (CH); For (INT I = 0; I <Len; I ++) {for (Int J = I; j-I <= 2 & J <Len; j ++) {If (CH [I] = '0 ') {DP [I + 1] + = DP [I]; break; // bounce} else {int TMP = char_to_int (I, j); If (TMP <= 255) DP [J + 1] = (DP [I] + dp [J + 1]) % 1234567; else break; // not satisfied, jump out} printf ("% d \ n", DP [Len] % 1234567);} return 0 ;}