给定一个整数数组arr [] ，任务是找到sum的所有值，以便对于sum [i]值，该数组可以划分为sum等于sum [i]的子数组。如果不能将array分成相等总和的子数组，则输出-1 。
例子：
Input: arr[] = {2, 2, 2, 1, 1, 2, 2}
Output: 2 4 6
The array can be divided into sub-arrays of sum 2, 4 and 6.
{2}, {2}, {2}, {1, 1}, {2} and {2}
{2, 2}, {2, 1, 1} and {2, 2}
{2, 2, 2} and {1, 1, 2, 2}
Input: arr[] = {1, 1, 2}
Output: 2
The array can be divided into sub-arrays of sum 2.
{1, 1} and {2}
方法：制作一个前缀和数组P [] ，其中P [i]存储从索引0到i的元素之和。总和S的除数只能是可能的子数组和。因此，对于每个除数，如果除数之和等于总和S的所有倍数都存在于数组P []中，那么那将是一个可能的子数组和。在地图中将P []的所有元素标记为1，以便查找。可以在sqrt(S)时间中检查所有除数。
下面是上述方法的实现：
C++
// C++ program to find the sums for which an array
// can be divided into subarrays of equal sum.
#include
using namespace std;
// Function to find the sums for which an array
// can be divided into subarrays of equal sum
void getSum(int a[], int n)
{
int P[n];
// Creating prefix sum array
P[0] = a[0];
for (int i = 1; i < n; i++)
P[i] = a[i] + P[i - 1];
// Total Sum
int S = P[n - 1];
// Initializing a Map for look-up
map hash;
// Setting all the present sum as 1
for (int i = 0; i < n; i++)
hash[P[i]] = 1;
// Set to store the subarray sum
set res;
// Check the divisors of S
for (int i = 1; i * i <= S; i++) {
if (S % i == 0) {
bool pres = true;
int div1 = i, div2 = S / i;
// Check if all multiples of div1 present or not
for (int j = div1; j <= S; j += div1) {
if (hash[j] != 1) {
pres = false;
break;
}
}
// If all multiples are present
if (pres and div1 != S)
res.insert(div1);
pres = true;
// Check if all multiples of div2 present or not
for (int j = S / i; j <= S; j += S / i) {
if (hash[j] != 1) {
pres = false;
break;
}
}
// If all multiples are present
if (pres and div2 != S)
res.insert(div2);
}
}
// If array cannot be divided into
// sub-arrays of equal sum
if(res.size() == 0) {
cout << "-1";
return;
}
// Printing the results
for (auto i : res)
cout << i << " ";
}
// Driver code
int main()
{
int a[] = { 1, 2, 1, 1, 1, 2, 1, 3 };
int n = sizeof(a) / sizeof(a[0]);
getSum(a, n);
return 0;
} Java
// Java program to find the sums for which an array
// can be divided into subarrays of equal sum.
import java.util.HashMap;
import java.util.HashSet;
class GFG {
// Function to find the sums for which an array
// can be divided into subarrays of equal sum
public static void getSum(int[] a, int n)
{
int[] P = new int[n];
// Creating prefix sum array
P[0] = a[0];
for (int i = 1; i < n; i++)
P[i] = a[i] + P[i - 1];
// Total Sum
int S = P[n - 1];
HashMap hash = new HashMap<>();
// Setting all the present sum as 1
for (int i = 0; i < n; i++)
hash.put(P[i], 1);
// Set to store the subarray sum
HashSet res = new HashSet<>();
// Check the divisors of S
for (int i = 1; i * i <= S; i++)
{
if (S % i == 0)
{
boolean pres = true;
int div1 = i, div2 = S / i;
// Check if all multiples of div1 present or not
for (int j = div1; j <= S; j += div1)
{
if (hash.get(j) == null || hash.get(j) != 1)
{
pres = false;
break;
}
}
// If all multiples are present
if (pres && div1 != S)
res.add(div1);
pres = true;
// Check if all multiples of div2 present or not
for (int j = S / i; j <= S; j += S / i)
{
if (hash.get(j) == null || hash.get(j) != 1)
{
pres = false;
break;
}
}
// If all multiples are present
if (pres && div2 != S)
res.add(div2);
}
}
// If array cannot be divided into
// sub-arrays of equal sum
if (res.size() == 0)
{
System.out.println("-1");
return;
}
// Printing the results
for (int i : res)
System.out.print(i + " ");
}
// Driver code
public static void main(String[] args)
{
int[] a = { 1, 2, 1, 1, 1, 2, 1, 3 };
int n = a.length;
getSum(a, n);
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 program to find the sums for
# which an array can be divided into
# subarrays of equal sum.
# from math lib import sqrt function
from math import sqrt
# Function to find the sums for which
# an array can be divided into subarrays
# of equal sum
def getSum(a, n) :
P = [0] * n
# Creating prefix sum array
P[0] = a[0]
for i in range(1, n) :
P[i] = a[i] + P[i - 1]
# Total Sum
S = P[n - 1]
# Initializing a Map for look-up
hash = {}
# Setting all the present sum as 1
for i in range(n) :
hash[P[i]] = 1
# Set to store the subarray sum
res = set()
# Check the divisors of S
for i in range(1, int(sqrt(S)) + 1) :
if (S % i == 0) :
pres = True;
div1 = i
div2 = S // i
# Check if all multiples of
# div1 present or not
for j in range(div1 , S + 1, div1) :
if j not in hash.keys() :
pres = False
break
# If all multiples are present
if (pres and div1 != S) :
res.add(div1)
pres = True
# Check if all multiples of div2
# present or not
for j in range(S // i , S + 1 , S // i) :
if j not in hash.keys():
pres = False;
break
# If all multiples are present
if (pres and div2 != S) :
res.add(div2)
# If array cannot be divided into
# sub-arrays of equal sum
if(len(res) == 0) :
print("-1")
return
# Printing the results
for i in res :
print(i, end = " ")
# Driver code
if __name__ == "__main__" :
a = [ 1, 2, 1, 1, 1, 2, 1, 3 ]
n = len(a)
getSum(a, n)
# This code is contributed by RyugaC#
// C# program to find the sums for which
// an array can be divided into subarrays
// of equal sum.
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the sums for which
// an array can be divided into subarrays
// of equal sum
public static void getSum(int[] a, int n)
{
int[] P = new int[n];
// Creating prefix sum array
P[0] = a[0];
for (int i = 1; i < n; i++)
P[i] = a[i] + P[i - 1];
// Total Sum
int S = P[n - 1];
Dictionary hash = new Dictionary();
// Setting all the present sum as 1
for (int i = 0; i < n; i++)
if(!hash.ContainsKey(P[i]))
hash.Add(P[i], 1);
// Set to store the subarray sum
HashSet res = new HashSet();
// Check the divisors of S
for (int i = 1; i * i <= S; i++)
{
if (S % i == 0)
{
Boolean pres = true;
int div1 = i, div2 = S / i;
// Check if all multiples of
// div1 present or not
for (int j = div1; j <= S; j += div1)
{
if (!hash.ContainsKey(j) ||
hash[j] != 1)
{
pres = false;
break;
}
}
// If all multiples are present
if (pres && div1 != S)
res.Add(div1);
pres = true;
// Check if all multiples of
// div2 present or not
for (int j = S / i;
j <= S; j += S / i)
{
if (hash[j] == 0 ||
hash[j] != 1)
{
pres = false;
break;
}
}
// If all multiples are present
if (pres && div2 != S)
res.Add(div2);
}
}
// If array cannot be divided into
// sub-arrays of equal sum
if (res.Count == 0)
{
Console.WriteLine("-1");
return;
}
// Printing the results
foreach (int i in res)
Console.Write(i + " ");
}
// Driver code
public static void Main(String[] args)
{
int[] a = { 1, 2, 1, 1, 1, 2, 1, 3 };
int n = a.Length;
getSum(a, n);
}
}
// This code is contributed by PrinciRaj1992 输出：
3 4 6
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