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 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
11640102497 1024
output
YES116401024
input
2842545891539281719112818110001009 1000
output
YES284254589153928171911281811000
input
12012 1
output
NO
又一次借鉴了袁学长的代码....
还是自己太菜了，啥都不咋会...
不过这次又学了好多....
比如怎么判断一个大数能否被一个数a整除
本题题意：就是去算出前一段能否整除a的同时能否使得后一段能整除b
思路：从前到后扫一遍记录前一段可以整除a的位置，从后到前扫一遍，如果后一段有可以整除b的位置，且前一段能整除a，就跳出循环，再输出，否则输出NO
判断一个大数num[1000010]能否被一个数a整除，代码：
#include <cstdio>#include <cstring>#include <iostream>using namespace std;int main(){ char num[1000010]; int a; while(scanf("%s", num)!=EOF) { scanf("%d", &a); int cur = 0; for(int i=0; i<strlen i cur num a if else cout return> <br> <br> <p></p> <p><strong>AC代码：</strong></p> <p><strong></strong></p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1000010;char num[maxn];bool jud[maxn];int main(){ while(scanf("%s", num)!=EOF) { int a, b; scanf("%d %d", &a, &b); memset(jud, 0, sizeof(jud)); //记录可以使前一段整除a的位置 int len = strlen(num); int cur = 0; for(int i=0; i<len i cur num a if jud bool ok="0;" int k="1," pos for>=0; i--) { cur+=(num[i]-'0')*k; cur%=b; if(cur==0 && jud[i-1] &&num[i]!='0') { ok = 1; pos = i; break; } k*=10; k%=b; } //输出答案 if(ok) { printf("YES\n"); for(int i=0; i<pos-1 i printf num for else return> <br> <br> <p></p> <p><br> </p> <p><br> </p> <p><br> </p> </pos-1></len></algorithm></cstring></cstdio>
