Problem - 409C - Codeforces(brute force + math)
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Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.
Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!
Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by a as a separate number, and the second (right) part is divisible by b as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values a and b.
Help Polycarpus and find any suitable method to cut the public key.
The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to106 digits. The second line contains a pair of space-separated positive integers a, b (1 ≤ a, b ≤ 108).
In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by a, and the right part must be divisible by b. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.
If there is no answer, print in a single line "NO" (without the quotes).
Sample test(s)
/* * author : qiu * coding : utf-8 * time : 2015-1-8 **/// code: 从左边扫一遍,求所有mod数,然后从右边扫一遍求mod数,比较输出ans!#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>#include <vector>#define debug puts("-----")#define pi (acos(-1.0))#define eps (1e-8)#define inf (1<<28)const int N=1000000+10;using namespace std;int ai[N];char key[N];int main(){ int a, b; while(cin >> key >> a >> b){ memset(ai, -1, sizeof(ai)); int len = strlen(key); int ff=0; for(int i=0; i<len-1; i++){ ff = (ff*10+(key[i]-'0')) %a; ai[i] = ff; } int tmp = 1; ff = 0; for(int i=len-1; i>0; i--){ ff = ((key[i]-'0')*tmp+ff) %b; tmp = (tmp*10) %b; if(ff==0 && !ai[i-1] && key[i]!='0') { cout << "YES" << endl; char temp=key[i]; key[i] = '\0'; cout << key << endl; key[i]=temp; cout << key+i << endl; return 0; } } cout << "NO" << endl; }return 0;}
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