原文:https://www . geeksforgeeks . org/除最小和最大元素之外的所有 k 大小子序列的乘积/
给定一个包含 n 元素和一个整数 k 的数组[]。任务是计算大小为 k 的子序列的所有元素的乘积，除了每个子序列的最小和最大元素。 注:由于答案可能会很大，所以将最终答案打印为 10 9 7 的 mod。
示例:
input : arr[] = {1, 2, 3 4}, k = 3
output : 36
*subsequences of length 3 are*:
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
excluding minimum and maximum elements from
each of the above subsequences, product will be:
*(2 * 2 * 3 * 3) = 36*.
input : arr[] = {10, 5, 16, 6}, k=3
output : 3600
天真方法:一个简单的方法是逐个生成所有可能的子序列，然后将除最大值和最小值之外的所有元素相乘，并进一步将所有元素相乘。因为总共会有【n】c【k】个子序列，所有子序列都有 k-2 个元素需要相乘，这是一项繁琐的工作。
高效方法:想法是首先对数组进行排序，因为我们是否考虑子序列或子集并不重要。 现在逐一统计每个元素的出现次数。 总的来说，一个数可以出现在(n-1)c(k-1)子序列中，其中(i)c(k-1)次作为最大元素出现，而(n-i-1)c(k-1)【t20 t21】次作为最小元素出现 因此，总共会出现元素:
(n-1)c(k-1) - (i)c(k-1) - (n-i-1)c(k-1) times. (let's say it x)
所以，首先我们要计算每个元素 a[i]的 x，然后乘以 a[i] x 倍。即()。 既然，对于大型阵列来说计算这个太难了，那么我们就用费马小定理。
下面是上述方法的实现:
c
// c program to find product of all
// subsequences of size k except the
// minimum and maximum elements
#include
using namespace std;
#define mod 1000000007
#define ll long long
#define max 101
// 2d array to store value of
// combinations ncr
ll c[max - 1][max - 1];
ll power(ll x, unsigned ll y)
{
unsigned ll res = 1;
x = x % mod;
while (y > 0) {
if (y & 1) {
res = (res * x) % mod;
}
y = y >> 1;
x = (x * x) % mod;
}
return res % mod;
}
// function to pre-calculate value of all
// combinations ncr
void combi(int n, int k)
{
int i, j;
for (i = 0; i <= n; i ) {
for (j = 0; j <= min(i, k); j ) {
if (j == 0 || j == i)
c[i][j] = 1;
else
c[i][j] = (c[i - 1][j - 1] % mod
c[i - 1][j] % mod) % mod;
}
}
}
// function to calculate product of all subsequences
// except the minimum and maximum elements
unsigned ll product(ll a[], int n, int k)
{
unsigned ll ans = 1;
// sorting array so that it becomes easy
// to calculate the number of times an
// element will come in first or last place
sort(a, a n);
// an element will occur 'powa' times in total
// of which 'powla' times it will be last element
// and 'powfa' times it will be first element
ll powa = c[n - 1][k - 1];
for (int i = 0; i < n; i ) {
ll powla = c[i][k - 1];
ll powfa = c[n - i - 1][k - 1];
// in total it will come
// powe = powa-powla-powfa times
ll powe = ((powa % mod) - (powla powfa) % mod mod) % mod;
// multiplying a[i] powe times using
// fermat little theorem under modulo
// mod for fast exponentiation
unsigned ll mul = power(a[i], powe) % mod;
ans = ((ans % mod) * (mul % mod)) % mod;
}
return ans % mod;
}
// driver code
int main()
{
// pre-calculation of all combinations
combi(100, 100);
ll arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof arr[0];
int k = 3;
unsigned ll ans = product(arr, n, k);
cout << ans << endl;
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to find product of all
// subsequences of size k except the
// minimum and maximum elements
import java.util.arrays;
class gfg
{
static int mod= 1000000007;
static int max =101;
// 2d array to store value of
// combinations ncr
static long c[][] = new long[max ][max];
static long power(long x, long y)
{
long res = 1;
x = x % mod;
while (y > 0)
{
if (y % 2== 1)
{
res = (res * x) % mod;
}
y = y >> 1;
x = (x * x) % mod;
}
return res % mod;
}
// function to pre-calculate value of all
// combinations ncr
static void combi(int n, int k)
{
int i, j;
for (i = 0; i <= n; i )
{
for (j = 0; j <= math.min(i, k); j )
{
if (j == 0 || j == i)
c[i][j] = 1;
else
c[i][j] = (c[i - 1][j - 1] % mod
c[i - 1][j] % mod) % mod;
}
}
}
// function to calculate product of all subsequences
// except the minimum and maximum elements
static long product(long a[], int n, int k)
{
long ans = 1;
// sorting array so that it becomes easy
// to calculate the number of times an
// element will come in first or last place
arrays.sort(a);
// an element will occur 'powa' times in total
// of which 'powla' times it will be last element
// and 'powfa' times it will be first element
long powa = c[n - 1][k - 1];
for (int i = 0; i < n; i )
{
long powla = c[i][k - 1];
long powfa = c[n - i - 1][k - 1];
// in total it will come
// powe = powa-powla-powfa times
long powe = ((powa % mod) - (powla powfa) % mod mod) % mod;
// multiplying a[i] powe times using
// fermat little theorem under modulo
// mod for fast exponentiation
long mul = power(a[i], powe) % mod;
ans = ((ans % mod) * (mul % mod)) % mod;
}
return ans % mod;
}
// driver code
public static void main(string[] args)
{
// pre-calculation of all combinations
combi(100, 100);
long arr[] = { 1, 2, 3, 4 };
int n = arr.length;
int k = 3;
long ans = product(arr, n, k);
system.out.println(ans);
}
}
/* this code contributed by princiraj1992 */
python 3
# python 3 program to find product of all
# subsequences of size k except the
# minimum and maximum elements
mod = 1000000007
max = 101
# 2d array to store value of
# combinations ncr
c = [[0 for i in range(max)] for j in range(max)]
def power(x,y):
res = 1
x = x % mod
while (y > 0):
if (y & 1):
res = (res * x) % mod
y = y >> 1
x = (x * x) % mod
return res % mod
# function to pre-calculate value of all
# combinations ncr
def combi(n, k):
for i in range(n 1):
for j in range(min(i, k) 1):
if (j == 0 or j == i):
c[i][j] = 1
else:
c[i][j] = (c[i - 1][j - 1] % mod
c[i - 1][j] % mod) % mod
# function to calculate product of all subsequences
# except the minimum and maximum elements
def product(a, n, k):
ans = 1
# sorting array so that it becomes easy
# to calculate the number of times an
# element will come in first or last place
a.sort(reverse = false)
# an element will occur 'powa' times in total
# of which 'powla' times it will be last element
# and 'powfa' times it will be first element
powa = c[n - 1][k - 1]
for i in range(n):
powla = c[i][k - 1]
powfa = c[n - i - 1][k - 1]
# in total it will come
# powe = powa-powla-powfa times
powe = ((powa % mod) - (powla powfa) % mod mod) % mod
# multiplying a[i] powe times using
# fermat little theorem under modulo
# mod for fast exponentiation
mul = power(a[i], powe) % mod
ans = ((ans % mod) * (mul % mod)) % mod
return ans % mod
# driver code
if __name__ == '__main__':
# pre-calculation of all combinations
combi(100, 100)
arr = [1, 2, 3, 4]
n = len(arr)
k = 3
ans = product(arr, n, k)
print(ans)
# this code is contributed by
# surendra_gangwar
c
// c# program to find product of all
// subsequences of size k except the
// minimum and maximum elements
using system;
class gfg
{
static int mod = 1000000007;
static int max = 101;
// 2d array to store value of
// combinations ncr
static long [,]c = new long[max, max];
static long power(long x, long y)
{
long res = 1;
x = x % mod;
while (y > 0)
{
if (y % 2 == 1)
{
res = (res * x) % mod;
}
y = y >> 1;
x = (x * x) % mod;
}
return res % mod;
}
// function to pre-calculate value
// of all combinations ncr
static void combi(int n, int k)
{
int i, j;
for (i = 0; i <= n; i )
{
for (j = 0;
j <= math.min(i, k); j )
{
if (j == 0 || j == i)
c[i, j] = 1;
else
c[i, j] = (c[i - 1, j - 1] % mod
c[i - 1, j] % mod) % mod;
}
}
}
// function to calculate product of
// all subsequences except
// the minimum and maximum elements
static long product(long []a, int n, int k)
{
long ans = 1;
// sorting array so that it becomes easy
// to calculate the number of times an
// element will come in first or last place
array.sort(a);
// an element will occur 'powa' times
// in total of which 'powla' times it
// will be last element and 'powfa' times
// it will be first element
long powa = c[n - 1, k - 1];
for (int i = 0; i < n; i )
{
long powla = c[i, k - 1];
long powfa = c[n - i - 1, k - 1];
// in total it will come
// powe = powa-powla-powfa times
long powe = ((powa % mod) -
(powla powfa) %
mod mod) % mod;
// multiplying a[i] powe times using
// fermat little theorem under modulo
// mod for fast exponentiation
long mul = power(a[i], powe) % mod;
ans = ((ans % mod) *
(mul % mod)) % mod;
}
return ans % mod;
}
// driver code
static public void main ()
{
// pre-calculation of all combinations
combi(100, 100);
long []arr = { 1, 2, 3, 4 };
int n = arr.length;
int k = 3;
long ans = product(arr, n, k);
console.writeline(ans);
}
}
// this code contributed by ajit
java 描述语言
output:
36
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