给定一个整数n和一个范围[l，r] ，任务是用给定范围内的整数找到大小为n的总子集的数量，以使其元素的总和可被3整除。
例子：
Input: n = 2, l = 1, r = 5
Output: 9
Possible sub-sets are {1, 2}, {2, 1}, {3, 3}, {5, 1}, {1, 5}, {4, 2}, {2, 4}, {5, 4} and {4, 5}
Input: n = 3, l = 9, r = 9
Output: 1
{9, 9, 9} is the only possible sub-set
方法：由于我们需要的子集元素的总和3.所以整除，而不是对数字关怀，我们将计算的数字，例如，他们给余数为0，1和2与3分别除以该公式如下：
For example, an element k such that k % 3 = 2 can be found as k = 3 * x + 2 for some integer x.
Then we have l ≤ (3 * x) + 2 ≤ r
l – 2 ≤ (3 * x) ≤ r – 2
ceil((l – 2) / 3) ≤ x ≤ floor((r – 2) / 3)
现在，通过动态编程dp [i] [j]，我们可以检查有多少元素将得出被3整除的总和。此处dp [i] [j]表示前i个元素的总和，这些元素除以3得到j 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
#define MOD 1000000007
#define ll long long int
using namespace std;
// Function to return the total number of
// required sub-sets
int totalSubSets(ll n, ll l, ll r)
{
// Variable to store total elements
// which on dividing by 3 give
// remainder 0, 1 and 2 respectively
ll zero = floor((double)r / 3)
- ceil((double)l / 3) + 1;
ll one = floor((double)(r - 1) / 3)
- ceil((double)(l - 1) / 3) + 1;
ll two = floor((double)(r - 2) / 3)
- ceil((double)(l - 2) / 3) + 1;
// Create a dp table
ll dp[n][3];
memset(dp, 0, sizeof(dp));
dp[0][0] = zero;
dp[0][1] = one;
dp[0][2] = two;
// Process for n states and store
// the sum (mod 3) for 0, 1 and 2
for (ll i = 1; i < n; ++i) {
// Use of MOD for large numbers
dp[i][0] = ((dp[i - 1][0] * zero)
+ (dp[i - 1][1] * two)
+ (dp[i - 1][2] * one))
% MOD;
dp[i][1] = ((dp[i - 1][0] * one)
+ (dp[i - 1][1] * zero)
+ (dp[i - 1][2] * two))
% MOD;
dp[i][2] = ((dp[i - 1][0] * two)
+ (dp[i - 1][1] * one)
+ (dp[i - 1][2] * zero))
% MOD;
}
// Final answer store at dp[n - 1][0]
return dp[n - 1][0];
}
// Driver Program
int main()
{
ll n = 5;
ll l = 10;
ll r = 100;
cout << totalSubSets(n, l, r);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static int MOD = 1000000007;
// Function to return the total number of
// required sub-sets
static int totalSubSets(int n, int l, int r)
{
// Variable to store total elements
// which on dividing by 3 give
// remainder 0, 1 and 2 respectively
int zero = (int)Math.floor((double)r / 3)
- (int)Math.ceil((double)l / 3) + 1;
int one = (int)Math.floor((double)(r - 1) / 3)
- (int)Math.ceil((double)(l - 1) / 3) + 1;
int two = (int)Math.floor((double)(r - 2) / 3)
- (int)Math.ceil((double)(l - 2) / 3) + 1;
// Create a dp table
int [][] dp = new int[n][3];
dp[0][0] = zero;
dp[0][1] = one;
dp[0][2] = two;
// Process for n states and store
// the sum (mod 3) for 0, 1 and 2
for (int i = 1; i < n; ++i)
{
// Use of MOD for large numbers
dp[i][0] = ((dp[i - 1][0] * zero)
+ (dp[i - 1][1] * two)
+ (dp[i - 1][2] * one))
% MOD;
dp[i][1] = ((dp[i - 1][0] * one)
+ (dp[i - 1][1] * zero)
+ (dp[i - 1][2] * two))
% MOD;
dp[i][2] = ((dp[i - 1][0] * two)
+ (dp[i - 1][1] * one)
+ (dp[i - 1][2] * zero))
% MOD;
}
// Final answer store at dp[n - 1][0]
return dp[n - 1][0];
}
// Driver Program
public static void main(String []args)
{
int n = 5;
int l = 10;
int r = 100;
System.out.println(totalSubSets(n, l, r));
}
}
// This code is contributed by ihritikPython3
# Python3 implementation of the approach
import math
# Function to return the total
# number of required sub-sets
def totalSubSets(n, l, r):
MOD = 1000000007 ;
# Variable to store total elements
# which on dividing by 3 give
# remainder 0, 1 and 2 respectively
zero = (math.floor(r / 3) -
math.ceil(l / 3) + 1);
one = (math.floor((r - 1) / 3) -
math.ceil((l - 1) / 3) + 1);
two = (math.floor((r - 2) / 3) -
math.ceil((l - 2) / 3) + 1);
# Create a dp table
dp = [[0 for x in range(3)]
for y in range(n)]
dp[0][0] = zero;
dp[0][1] = one;
dp[0][2] = two;
# Process for n states and store
# the sum (mod 3) for 0, 1 and 2
for i in range(1, n):
# Use of MOD for large numbers
dp[i][0] = ((dp[i - 1][0] * zero) +
(dp[i - 1][1] * two) +
(dp[i - 1][2] * one)) % MOD;
dp[i][1] = ((dp[i - 1][0] * one) +
(dp[i - 1][1] * zero) +
(dp[i - 1][2] * two)) % MOD;
dp[i][2] = ((dp[i - 1][0] * two)+
(dp[i - 1][1] * one) +
(dp[i - 1][2] * zero)) % MOD;
# Final answer store at dp[n - 1][0]
return dp[n - 1][0];
# Driver Code
n = 5;
l = 10;
r = 100;
print(totalSubSets(n, l, r));
# This code is contributed
# by chandan_jnuC#
// C# implementation of the approach
using System;
class GFG
{
static int MOD = 1000000007;
// Function to return the total number of
// required sub-sets
static int totalSubSets(int n, int l, int r)
{
// Variable to store total elements
// which on dividing by 3 give
// remainder 0, 1 and 2 respectively
int zero = (int)Math.Floor((double)r / 3)
- (int)Math.Ceiling((double)l / 3) + 1;
int one = (int)Math.Floor((double)(r - 1) / 3)
- (int)Math.Ceiling((double)(l - 1) / 3) + 1;
int two = (int)Math.Floor((double)(r - 2) / 3)
- (int)Math.Ceiling((double)(l - 2) / 3) + 1;
// Create a dp table
int [, ] dp = new int[n, 3];
dp[0,0] = zero;
dp[0,1] = one;
dp[0,2] = two;
// Process for n states and store
// the sum (mod 3) for 0, 1 and 2
for (int i = 1; i < n; ++i)
{
// Use of MOD for large numbers
dp[i,0] = ((dp[i - 1, 0] * zero)
+ (dp[i - 1, 1] * two)
+ (dp[i - 1, 2] * one))
% MOD;
dp[i,1] = ((dp[i - 1, 0] * one)
+ (dp[i - 1, 1] * zero)
+ (dp[i - 1, 2] * two))
% MOD;
dp[i,2] = ((dp[i - 1, 0] * two)
+ (dp[i - 1, 1] * one)
+ (dp[i - 1, 2] * zero))
% MOD;
}
// Final answer store at dp[n - 1,0]
return dp[n - 1, 0];
}
// Driver Program
public static void Main()
{
int n = 5;
int l = 10;
int r = 100;
Console.WriteLine(totalSubSets(n, l, r));
}
}
// This code is contributed by ihritikPHP
输出：
80107136