【Codeforces 490C】Hacking Cypher
【链接】 我是链接,点我呀:)
【题意】
【题解】
枚举分割的断点i 枚举的时候用同余率算出来s[1..i]和a以及b取余的结果 怎么得到s[i+1..len-1]呢? 只要用s[1..len]-s[1..b]就可以了 乘的时候可能会爆int,小心处理【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)1e6;
static class Task{
String s;
long a,b;
long sumb = 0,prea = 0,preb = 0;
long aftb[] = new long[N+10];
public void solve(InputReader in,PrintWriter out) {
s = in.next();
a = in.nextInt();b = in.nextInt();
aftb[(int)s.length()] = 1%b;
for (int i = (int)s.length()-1;i>=0;i--) {
aftb[i] = aftb[i+1]*10%b;
}
for (int i = 0;i < (int)s.length();i++) {
sumb = sumb*10 + s.charAt(i)-'0';
sumb = sumb%b;
}
for (int i = 0;i < (int)s.length()-1;i++) {
prea = prea * 10 + s.charAt(i)-'0';
prea = prea%a;
preb = preb * 10 + s.charAt(i)-'0';
preb = preb%b;
long temp = (sumb-preb*aftb[i+1]%b)%b;
if (temp<0) temp+=b;
if (temp==0 && prea ==0) {
if (s.charAt(i+1)=='0') continue;
out.println("YES");
for (int j = 0;j <=i;j++) {
out.print(s.charAt(j));
}
out.println();
for (int j = i+1;j < (int)s.length();j++) {
out.print(s.charAt(j));
}
return;
}
}
out.println("NO");
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}