Hacking Cypher（数学）

C. Hacking Cypher

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 to 106 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

     题意：

     给出一个长度为 n 的数字，后给出 a 和 b，问能否把这个数字分成两段，使之前半段能整除 a，后半段能整除 b。

 

     思路：

     数学。直接暴搜无疑会 TLE，从前面往后面扫的时候用 tap 标记这一位能否被 a 整除，边扫边还可以边计算能否被整除。之后从后面开始往前扫，但是从后面往前扫判断能否被整除就要看下公式了。因为

     一个数 abc = （a X 100 + b X 10 + c），所以当模的时候就等于 a % M X 100 % M + b % M X 10 % M + c % M，所以从后面往前面扫的时候记录 10 的次方模就好，之后根据公式就可以算出了。

     如果这位能被整除，且这一位的前一位 tap 值等于 1，则 break，输出两段数字即可。

 

     AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int MAX = 1000005;

char num[MAX];

int tap[MAX];

int main() {
    int a, b, len;

    memset(tap, 0, sizeof(tap));

    scanf("%s", num);
    scanf("%d%d", &a, &b);

    len = strlen(num);

    int ans = 0;
    for (int i = 0; i < len; ++i) {
        ans *= 10;
        ans += num[i] - '0';
        ans %= a;
        if (!ans) tap[i] = 1;
    }

    int temp = 0, in = 1;
    ans = 0;
    for (int i = len - 1; i >= 0; --i) {
        ans = (in * (num[i] - '0')) % b + ans;
        ans %= b;
        in = (in * 10) % b;

        if (!ans && num[i] != '0' && tap[i - 1]) {
            printf("YES\n");
            temp = 1;
            for (int j = 0; j <= i - 1; ++j) {
                printf("%c", num[j]);
            }
            printf("\n");
            for (int j = i; j < len; ++j) {
                printf("%c", num[j]);
            }
            printf("\n");
            break;
        }
    }

    if (!temp) printf("NO\n");

    return 0;
}