【链接】 我是链接,点我呀:)
【题意】
让你把数组分成3个连续的部分
每个部分的和要一样
问你有多少种分法
【题解】
先处理出来num[i]
表示i..n这里面有多少个j
满足aft[j] = aft[i]/2
这aft[i]=a[j]+a[j+1]..+a[n]
然后for从1..n
看看pre[i]*2=aft[i+1]是否成立。
如果成立的话那么答案就加上num[i+1]
【代码】
import java.io.*;
import java.util.*;
public class Main {
static int N = (int)5e5;
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
in = new InputReader();
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static class Task{
public void solve(InputReader in,PrintWriter out) {
int n = in.nextInt();
int []a = new int[N+10];
long []pre = new long[N+10];
long []aft = new long[N+10];
int []num = new int[N+10];
for (int i = 1;i <= n;i++) a[i] = in.nextInt();
for (int i = 1;i <= n;i++) pre[i] = pre[i-1]+a[i];
for (int i = n;i >= 1;i--) aft[i] = aft[i+1]+a[i];
Hashtable<Long, Integer> dic = new Hashtable<Long,Integer>();
for (int i = n;i >= 1;i--) {
if (aft[i]%2==0) {
long temp = aft[i]/2;
if (dic.get(temp)!=null)
num[i] = dic.get(temp);
}
Integer temp1 = dic.get(aft[i]);
if (temp1==null) {
dic.put( aft[i], 1);
}else {
dic.put(aft[i], temp1+1);
}
}
long ans = 0;
for (int i = 1;i <=n;i++) {
long cur = pre[i];
cur = pre[n]-cur;
if (cur!=pre[i]*2) {
continue;
}
ans = ans + num[i+1];
}
out.println(ans);
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader() {
br = new BufferedReader(new InputStreamReader(System.in));
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());
}
}
}
原文地址:https://www.cnblogs.com/AWCXV/p/10359024.html
时间: 2024-10-25 17:21:42