Bestcoder #49

1001 Untitled

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

/**
* 2015年8月1日 下午7:15:14
* PrjName:Bc49-01
* @ Semprathlon
*/
import java.io.*;
import java.util.*;
public class Main {
static ArrayList<Integer> v=new ArrayList<Integer>();
static int[] a,f;
static int lowbit(int x){
return x&(-x);
}
static int get(int x){
int res=0;
while(x>0){
res++;
x-=lowbit(x);
}
return res;
}
public static void main(String[] args) throws IOException{
// TODO Auto-generated method stub
InputReader in=new InputReader(System.in);
PrintWriter out=new PrintWriter(System.out);
int T=in.nextInt();
while(T-->0){
v.clear();
int n=in.nextInt();
int m=in.nextInt();
for(int i=1;i<=n;i++)
v.add(in.nextInt());
Collections.sort(v);
f=new int[1<<n];
Arrays.fill(f, -1);
f[0]=m;
//out.println(n);
int ans=Integer.MAX_VALUE;
for(int i=0;i<(1<<n);i++)
if (f[i]>0)
for(int j=0;j<n;j++)
if (((1<<j)&i)==0){
int k=i|(1<<j);
f[k]=f[i]%v.get(j);
if (f[k]==0)
ans=Math.min(ans, get(k));
}
if (ans==Integer.MAX_VALUE) out.println(-1);
else out.println(ans);
}
out.flush();
out.close();
}
}

1002 Three Palindromes

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Can we divided a given string S into three nonempty palindromes?

Input

First line contains a single integer latexT≤20latex T \leq 20 which denotes the number of test cases.

For each test case , there is an single line contains a string S which only consist of lowercase English letters.latex1≤|s|≤20000latex 1\leq |s| \leq 20000

Output

For each case, output the “Yes” or “No” in a single line.

Sample Input

2

abc

abaadada

Sample Output

Yes

No

赛时的Hash做法：

/**
* 2015年8月1日 下午7:47:15
* PrjName:Bc49-02
* @ Semprathlon
*/
import java.io.*;
import java.util.*;
public class Main {
static long pri=29L;
static long lh(String s,long res,int p){
return res*pri+(long)(s.charAt(p)-'a');
}
static long rh(String s,long res,int p,long m){
return res+(long)(s.charAt(p)-'a')*m;
}
static BitSet v=new BitSet(20010);
public static void main(String[] args) throws IOException{
// TODO Auto-generated method stub
InputReader in=new InputReader(System.in);
PrintWriter out=new PrintWriter(System.out);
int T=in.nextInt();
while(T-->0){
String s=new String(in.next());
int len=s.length();
//s=s+"#"+(new StringBuffer(s).reverse().toString());
//out.println(s);
boolean ans=false;
long l1=0L,r1=0L,tmp=1L;
for(int i=0;i<len;i++){
if (ans) break;
l1=lh(s, l1, i);
r1=rh(s, r1, i, tmp);
//out.println(i+" "+l1+" "+r1);
tmp*=pri;
//if ((i&1)>0) continue;
if (l1==r1){
//out.println("f"+i);
long l2=0L,r2=0L,tmp2=1L;
v.clear();
for(int j=i+1;j<len;j++){
l2=lh(s,l2,j);
r2=rh(s,r2,j,tmp2);
tmp2*=pri;
//if (((j-i+1)&1)>0) continue;
if (l2==r2) v.set(j);
}

l2=0L;r2=0L;tmp2=1L;
for(int j=len-1;j>i;j--){
l2=lh(s,l2,j);
r2=rh(s,r2,j,tmp2);
tmp2*=pri;
//if (((len-j+1)&1)>0) continue;
if (l2==r2&&v.get(j-1)){
ans=true;break;
}
}
}
}
out.println(ans?"Yes":"No");
}
out.flush();
out.close();
}
}

首次尝试去hack别人不成，还被人黑了，立马TLE

题解介绍的多种方法中，个人认为二分+hash代码量较少，如何实现？