给定一个整数k和一个数组arr [] ，任务是计算总和等于k的某个正整数幂的子数组的数量。
例子：
Input: arr[] = { 2, 2, 2, 2 } K = 2
Output: 8
Sub-arrays with below indexes are valid:
[1, 1], [2, 2], [3, 3], [4, 4], [1, 2],
[2, 3], [3, 4], [1, 4]
Input: arr[] = { 3, -6, -3, 12 } K = -3
Output: 3
天真的方法：天真的方法是遍历所有子数组，并检查每个子数组的总和是否等于k的某个整数幂。
高效的方法：更好的方法是维护一个前缀和数组prefix_sum和一个映射m ，该映射m将前缀和映射到其计数。 m [a] = 1表示a是某个前缀的前缀和。
Iterate through the array in the backward direction and carefully follow the below discussion. Consider that while traversing the array we are at the ith index and after traversing an index we perform the operation op = m[prefix_sum[i]]++. Hence, when we are at index i, op hasn’t been performed yet. See the code for a vivid explanation.
If m[a + b] = c where a = prefix_sum[i], b = kp and c is the value fetched from the map, then it means that starting from the ith index to the end of the array, there are c sub-arrays whose sum is equal to b. Add c to the current sum.
This is because for every index j > i, m[prefix_sum[j]]++ has been performed. Therefore, the map has information about prefix sums of prefixes ending at j > i. On adding b to the prefix sum we can get the count of all those sums a + b which will indicate that a sub-array exists that has sum equal to b.
注意：k = 1和k = -1需要分开处理。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
#define ll long long
#define MAX 100005
using namespace std;
// Function to count number of sub-arrays
// whose sum is k^p where p>=0
ll countSubarrays(int* arr, int n, int k)
{
ll prefix_sum[MAX];
prefix_sum[0] = 0;
partial_sum(arr, arr + n, prefix_sum + 1);
ll sum;
if (k == 1) {
sum = 0;
map m;
for (int i = n; i >= 0; i--) {
// If m[a+b] = c, then add c to the current sum.
if (m.find(prefix_sum[i] + 1) != m.end())
sum += m[prefix_sum[i] + 1];
// Increase count of prefix sum.
m[prefix_sum[i]]++;
}
return sum;
}
if (k == -1) {
sum = 0;
map m;
for (int i = n; i >= 0; i--) {
// If m[a+b] = c, then add c to the current sum.
if (m.find(prefix_sum[i] + 1) != m.end())
sum += m[prefix_sum[i] + 1];
if (m.find(prefix_sum[i] - 1) != m.end())
sum += m[prefix_sum[i] - 1];
// Increase count of prefix sum.
m[prefix_sum[i]]++;
}
return sum;
}
sum = 0;
// b = k^p, p>=0
ll b;
map m;
for (int i = n; i >= 0; i--) {
b = 1;
while (true) {
// k^m can be maximum equal to 10^14.
if (b > 100000000000000)
break;
// If m[a+b] = c, then add c to the current sum.
if (m.find(prefix_sum[i] + b) != m.end())
sum += m[prefix_sum[i] + b];
b *= k;
}
// Increase count of prefix sum.
m[prefix_sum[i]]++;
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 2, 2, 2, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
cout << countSubarrays(arr, n, k);
return 0;
} Java
// Java implementation of the
// above approach
import java.util.*;
class GFG{
static final int MAX = 100005;
// partial_sum
static long[] partial_sum(long []prefix_sum,
int[]arr, int n)
{
for (int i = 1; i <= n; i++)
{
prefix_sum[i] = (prefix_sum[i - 1] +
arr[i - 1]);
}
return prefix_sum;
}
// Function to count number of
// sub-arrays whose sum is k^p
// where p>=0
static int countSubarrays(int []arr,
int n, int k)
{
long []prefix_sum = new long[MAX];
prefix_sum[0] = 0;
prefix_sum = partial_sum(prefix_sum ,
arr, n);
int sum;
if (k == 1)
{
sum = 0;
HashMap m = new HashMap<>();
for (int i = n; i >= 0; i--)
{
// If m[a+b] = c, then add c to
// the current sum.
if (m.containsKey(prefix_sum[i] + 1))
sum += m.get(prefix_sum[i] + 1);
// Increase count of prefix sum.
if(m.containsKey(prefix_sum[i]))
m.put(prefix_sum[i],
m.get(prefix_sum[i]) + 1);
else
m.put(prefix_sum[i], 1);
}
return sum;
}
if (k == -1)
{
sum = 0;
HashMap m = new HashMap<>();
for (int i = n; i >= 0; i--)
{
// If m[a+b] = c, then add c to
// the current sum.
if (m.containsKey(prefix_sum[i] + 1))
sum += m.get(prefix_sum[i] + 1);
if (m.containsKey(prefix_sum[i] - 1))
sum += m.get(prefix_sum[i] - 1);
// Increase count of prefix sum.
if(m.containsKey(prefix_sum[i]))
m.put(prefix_sum[i],
m.get(prefix_sum[i]) + 1);
else
m.put(prefix_sum[i], 1);
}
return sum;
}
sum = 0;
// b = k^p, p>=0
long b, l = 100000000000000L;
HashMap m = new HashMap<>();
for (int i = n; i >= 0; i--)
{
b = 1;
while (true)
{
// k^m can be maximum equal
// to 10^14.
if (b > l)
break;
// If m[a+b] = c, then add c to
// the current sum.
if (m.containsKey(prefix_sum[i] + b))
sum += m.get(prefix_sum[i] + b);
b *= k;
}
// Increase count of prefix sum.
if(m.containsKey(prefix_sum[i]))
m.put((prefix_sum[i]),
m.get(prefix_sum[i]) + 1);
else
m.put((prefix_sum[i]), 1);
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {2, 2, 2, 2};
int n = arr.length;
int k = 2;
System.out.print(countSubarrays(arr,
n, k));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of
# the above approach
from collections import defaultdict
MAX = 100005
def partial_sum(prefix_sum,
arr, n):
for i in range(1 , n + 1):
prefix_sum[i] = (prefix_sum[i - 1] +
arr[i - 1])
return prefix_sum
# Function to count number of
# sub-arrays whose sum is k^p
# where p>=0
def countSubarrays(arr, n, k):
prefix_sum = [0] * MAX
prefix_sum[0] = 0
prefix_sum = partial_sum(prefix_sum,
arr, n)
if (k == 1):
sum = 0
m = defaultdict(int)
for i in range(n, -1, -1):
# If m[a+b] = c, then add
# c to the current sum.
if ((prefix_sum[i] + 1) in m):
sum += m[prefix_sum[i] + 1]
# Increase count of prefix sum.
m[prefix_sum[i]] += 1
return sum
if (k == -1):
sum = 0
m = defaultdict(int)
for i in range(n, -1, -1):
# If m[a+b] = c, then add c
# to the current sum.
if ((prefix_sum[i] + 1) in m):
sum += m[prefix_sum[i] + 1]
if ((prefix_sum[i] - 1) in m):
sum += m[prefix_sum[i] - 1]
# Increase count of prefix sum.
m[prefix_sum[i]] += 1
return sum
sum = 0
# b = k^p, p>=0
m = defaultdict(int)
for i in range(n, -1, -1):
b = 1
while (True):
# k^m can be maximum equal
# to 10^14.
if (b > 100000000000000):
break
# If m[a+b] = c, then add c
# to the current sum.
if ((prefix_sum[i] + b) in m):
sum += m[prefix_sum[i] + b]
b *= k
# Increase count of prefix
# sum.
m[prefix_sum[i]] += 1
return sum
# Driver code
if __name__ == "__main__":
arr = [2, 2, 2, 2]
n = len(arr)
k = 2
print(countSubarrays(arr, n, k))
# This code is contributed by ChitranayalC#
// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
class GFG{
static readonly int MAX = 100005;
// partial_sum
static long[] partial_sum(long []prefix_sum,
int[]arr, int n)
{
for (int i = 1; i <= n; i++)
{
prefix_sum[i] = (prefix_sum[i - 1] +
arr[i - 1]);
}
return prefix_sum;
}
// Function to count number of
// sub-arrays whose sum is k^p
// where p>=0
static int countSubarrays(int []arr,
int n, int k)
{
long []prefix_sum = new long[MAX];
prefix_sum[0] = 0;
prefix_sum = partial_sum(prefix_sum ,
arr, n);
int sum;
if (k == 1)
{
sum = 0;
Dictionary mp =
new Dictionary();
for (int i = n; i >= 0; i--)
{
// If m[a+b] = c, then add c to
// the current sum.
if (mp.ContainsKey(prefix_sum[i] + 1))
sum += mp[prefix_sum[i] + 1];
// Increase count of prefix sum.
if(mp.ContainsKey(prefix_sum[i]))
mp.Add(prefix_sum[i],
mp[prefix_sum[i]] + 1);
else
mp.Add(prefix_sum[i], 1);
}
return sum;
}
if (k == -1)
{
sum = 0;
Dictionary map =
new Dictionary();
for (int i = n; i >= 0; i--)
{
// If m[a+b] = c, then add c to
// the current sum.
if (map.ContainsKey(prefix_sum[i] + 1))
sum += map[prefix_sum[i] + 1];
if (map.ContainsKey(prefix_sum[i] - 1))
sum += map[prefix_sum[i] - 1];
// Increase count of prefix sum.
if(map.ContainsKey(prefix_sum[i]))
map.Add(prefix_sum[i],
map[prefix_sum[i]] + 1);
else
map.Add(prefix_sum[i], 1);
}
return sum;
}
sum = 0;
// b = k^p, p>=0
long b, l = 100000000000000L;
Dictionary m =
new Dictionary();
for (int i = n; i >= 0; i--)
{
b = 1;
while (true)
{
// k^m can be maximum equal
// to 10^14.
if (b > l)
break;
// If m[a+b] = c, then add c to
// the current sum.
if (m.ContainsKey(prefix_sum[i] + b))
sum += m[prefix_sum[i] + b];
b *= k;
}
// Increase count of prefix sum.
if(m.ContainsKey(prefix_sum[i]))
m.Add((prefix_sum[i]),
m[prefix_sum[i]] + 1);
else
m.Add((prefix_sum[i]), 1);
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {2, 2, 2, 2};
int n = arr.Length;
int k = 2;
Console.Write(countSubarrays(arr,
n, k));
}
}
// This code is contributed by shikhasingrajput 输出：
8
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