Digital problems (c ++ implementation), digital

Digital problems (c ++ implementation), digital

Description: The LCD digital tube uses ten numbers represented by seven numbers. The horizontal and vertical short strokes are called one stroke, that is, there are three strokes in 7 and seven strokes in 8. 10 digits must be arranged.
Two Adjacent numbers can both be composed of another number plus or minus a few, but cannot be added or subtracted. For example, 7 → 3 is allowed, and 7 → 2 is not allowed. Enter a group of numbers to determine whether the group meets the preceding rules.

Input: enter a value ranging from 0 ~ 9. Separate numbers with spaces and end with-1 as the input.

Output: Enter YES or NO.

Input:

４１０７３９５６８２ 3  5 1 6  2  7 9  0  4  80
output：

YESNO

Analysis: there is a way to order each table so that each number can be expressed as a string, for example, 1 is "35 ", then, determine whether the adjacent two match the rule. You can use a character string with a larger number segment to determine whether it is a subset of a larger string.

1 # include <iostream> 2 # include <string> 3 using namespace std; 4 5 // determine whether the adjacent two can convert each other to 6 int s (string max, string min) 7 {8 int n = 0; 9 for (int I = 0; I <min. size (); I ++) 10 {11 for (int j = 0; j <max. size (); j ++) 12 {13 if (min [I] = max [j]) 14 n ++; 15} 16} 17 if (n = min. size () 18 return 1; 19 else20 return 0; 21} 22 23 // judgment function 24 int cal (int a []) 25 {26 int k; 27 string B [10] = {"123567", "35", "23467 ", "23456", "1345", "12456", "124567", "235", "1234567", "123456 "}; // mark the 7-digit number in the digital tube with numbers 1-7 and store each number with strings 28 // compare the adjacent two 29 for (int I = 0; I <9; I ++) 30 {31 32 if (B [a [I]. size ()> B [a [I + 1]. size () 33 k = s (B [a [I], B [a [I + 1]); 34 else35 k = s (B [a [I + 1], B [a [I]); 36 if (! K) 37 return 0; 38} 39 return 1; 40} 41 42 int main () 43 {44 int a [10]; 45 while (cin> a [0]) & (a [0]! =-1) 46 {47 for (int I = 1; I <10; I ++) 48 cin> a [I]; 49 if (cal ()) 50 cout <"Yes" <endl; 51 else52 cout <"No" <endl; 53} 54 system ("pause"); 55 return 0; 56}