解题思路
- 发现正着决策是否划分会导致乘数增加,影响到后面,可能使结果变劣
- 而倒着决策,当前选择划分使乘数增加,但后面的一定进行了最优决策,乘数增加不会使结果变劣
- 考虑什么时候会划分
- 只有当当前累加和大于0会进行划分
- 考虑将乘划分数转为累加和,等于加了次
- 倒着开始,保证每个数的累加次数非递减,若,则加入答案
- 注意处理最终
java
import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.Vector;
//implements Runnable
public class Main {
static long md=(long)1e9+7;
static long Linf=Long.MAX_VALUE/2;
static int inf=Integer.MAX_VALUE/2;
static int N=200010;
static int n=0;
static int m=0;
static void solve() throws Exception{
AReader input=new AReader();
// Scanner input=new Scanner(System.in);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
String al="abcdefghijklmnopqrstuvwxyz";
int T=input.nextInt();
while(T>0) {
T--;
n=input.nextInt();
long[] a=new long[n+1];
for(int i=1;i<=n;++i)a[i]=input.nextLong();
long ans=0;
long sum=0;
for(int i=n;i>=1;--i) {
sum+=a[i];
if(sum>0) {
ans+=sum;
}
}
if(sum<0)ans+=sum;
out.println(ans);
}
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
solve();
}
// public static final void main(String[] args) throws Exception {
// new Thread(null, new Tx2(), "线程名字", 1 << 27).start();
// }
// @Override
// public void run() {
// try {
// //原本main函数的内容
// solve();
//
// } catch (Exception e) {
// }
// }
static
class AReader{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public AReader(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
//确定下一个token只有一个字符的时候再用
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public byte nextByte() throws IOException{
return Byte.parseByte(next());
}
public short nextShort() throws IOException{
return Short.parseShort(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public void println() throws IOException {
bw.newLine();
}
public void println(int[] arr) throws IOException{
for (int value : arr) {
bw.write(value + " ");
}
println();
}
public void println(int l, int r, int[] arr) throws IOException{
for (int i = l; i <= r; i ++) {
bw.write(arr[i] + " ");
}
println();
}
public void println(int a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
}
public void print(int a) throws IOException{
bw.write(String.valueOf(a));
}
public void println(String a) throws IOException{
bw.write(a);
bw.newLine();
}
public void print(String a) throws IOException{
bw.write(a);
}
public void println(long a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
}
public void print(long a) throws IOException{
bw.write(String.valueOf(a));
}
public void println(double a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
}
public void print(double a) throws IOException{
bw.write(String.valueOf(a));
}
public void print(char a) throws IOException{
bw.write(String.valueOf(a));
}
public void println(char a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
}
}
}