Codeforces 490C Hacking Cypher-白红宇

Codeforces 490C Hacking Cypher

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发布时间：2019-06-29

本文共 3670 字，大约阅读时间需要 12 分钟。

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.

Input

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).

Output

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)

input

116401024 97 1024

output

YES 11640 1024

input

284254589153928171911281811000 1009 1000

output

YES 2842545891539 28171911281811000

input

120 12 1

output

NO 给出一个长度为1e6的数你 , 然后两个数 a , b 最大是1e8 。 问能否把 第一个数切成2半 ，然后半部分整除 a  , 后半部分整除 b 。 想了挺久 。 一开始以为要大数。 其实不然 。 对于一个数 a , 在最后加一个个位数 b 的话 ： ( a*10 + b ) % mod = ( ( a % mod ) * ( 10 % mod ) % mod + b % mod ) % mod ; 然而 , 在最前面加的话 。 按上面的公式可以求出 10^i % mod 的值， 所以又知道 ( b * 10^i ) % mod 的值 。 那么 （ b*10^i + a ）% mod  又可以求出来了。 若果余数是0的话就可以整除了 。 在判断一下有没有前缀0即可

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                    using namespace std;typedef long long LL;const int N = 1000010;const int inf = 1e7+7;const double PI = acos(-1.0);const double eps = 1e-6 ;#define base 10000char str[N];int num[N] ;LL a , b , len ;bool can1[N] , can2[N];void run() { for( int i = 0 ; i + 1 < len ; ++i ) if( can1[i] && can2[i+1] && num[i+1] ){ cout << "YES" << endl ; for( int j = 0 ; j <= i ; ++j ) cout << str[j] ; cout << endl ; for( int j = i + 1 ; j < len ; ++j ) cout << str[j] ; cout << endl ; return ; } cout << "NO" << endl ;}int main(){ #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL ios::sync_with_stdio(false); while( scanf("%s",str) != EOF ) { scanf("%d%d",&a,&b); memset( can1 , false , sizeof can1 ); memset( can2 , false , sizeof can2 ); len = strlen( str ); for( int i = 0 ; i < len ; ++i ) num[i] = str[i] - '0' ; LL num1 = 0 ; for( int i = 0 ; i < len ; ++i ){ num1 = ( ( ( num1 % a ) * ( 10 % a ) ) % a + num[i] % a ) % a ; if( !num1 ) can1[i] = true ; } LL num2 = 0 , num3 = 1 ; for( int i = len -1 ; i >= 0 ; --i ){ LL temp = ( num3 * num[i] ) % b ; num2 = ( temp + num2 ) % b ; if( !num2 ) can2[i] = true ; num3 = num3 * ( 10 % b ) % b ; } run(); }}

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转载于:https://www.cnblogs.com/hlmark/p/4117544.html

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