Codeforces 490C. Hacking Cypher Problem Solving report

Title Link: Http://codeforces.com/problemset/problem/490/C

Title meaning: Give a string that may have a 10^6 bit length and no leading 0 integer, ask whether it can be divided into two, so that the previous part is divisible by a and the latter part is divisible by B, you can output "YES" and the division after the second part, otherwise output "NO".

See the string so long, suddenly scared the ~ ~ ~ with mod! It depends on the accumulation of mathematics ...

The usual way to do it--the brute force enumeration, of course, cannot be omitted, so from left to right until the penultimate number, the result of MoD A is calculated and saved to the vis[] array (Vis is initialized to-1). Then from right forward until the second number of the number of CIS, the results of the MoD B, when the result is 0 (that is, can be divisible by B), and vis[before a digit] = 0 (indicating that the preceding number can be divisible by a), there is a simple omission, there is no leading 0 (s[i in the corresponding code]! = Then the representative finds a solution. If you don't have any answers after the traversal, the output "no"

1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 using namespace std;
 6
 7 const int maxn = 1e6 + 5;
 8 int vis[maxn]; // Save the result of mod a of 0~len-1 numbers in string
 9 string s;
10
11 int main()
12 {
13 #ifndef ONLINE_JUDGE
14 freopen("in.txt", "r", stdin);
15 #endif // ONLINE_JUDGE
16 int a, b;
17 while (cin >> s >> a >> b)
18 {
19 memset(vis, -1, sizeof(vis));
20 int tmp = 0;
21 int len = s.size();
twenty two        
23 for (int i = 0; i <len-1; i++) // At least one must be left for b
twenty four         {
25 tmp = (tmp * 10 + (s[i]-‘0’))% a; // Some small knowledge of mathematics
26 vis[i] = tmp;
27}
28
29 int base = 1;
30 tmp = 0;
31 for (int i = len-1; i> 0; i--) // Similarly, a must have at least one digit
32 {
33 tmp = ((s[i]-‘0‘) * base + tmp)% b;
34 base = base * 10% b;
35 if (tmp == 0 && !vis[i-1] && s[i] != ‘0’)
36 {
37 cout << "YES" << endl;
38 cout << s.substr(0, i) << endl;
39 cout << s.substr(i, len-i) << endl;
40 return 0;
41}
42}
43 cout << "NO" << endl;
44}
45 return 0;
46} 

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Codeforces 490C. Hacking Cypher Problem Solving report