计数方法使由 K 连接形成的数字字符串可被 5 整除
给定一个由N个数字和一个整数K组成的字符串S ，任务是计算从连接字符串S形成的数字中删除数字的次数， K次，使得结果字符串可以被5整除.由于计数可能非常大，因此将其打印为模10 9 + 7 。
例子：
Input: S = 1256, K = 1
Output: 4
Explanation:
Following are the ways to remove the characters so that the string S(= “1256”) is divisible by 5:
- Remove the character 6 from the string S modifies the string to 125, which is divisible by 5.
- Remove the character 1, and 6 from the string S modifies the string to 25, which is divisible by 5.
- Remove the character 2, and 6 from the string S modifies the string to 15, which is divisible by 5.
- Remove the character 1, 2, and 6 from the string S modifies the string to 5, which is divisible by 5.
Therefore, the total count of numbers formed which is divisible by 5 is 4.
Input: S = 13390, K = 2
Output: 528
方法：给定的问题可以通过以下事实来解决：当且仅当它的最后一位数字是0或5时，该数字可以被 5 整除。如果sol[i]是形成以i结尾的可被5整除的数字的方式数，则数字计数由(sol[1] + sol[2] + … + sol[N])给出。对于每个索引i ，在 在i的右边，选择删除所有数字，在i的左边，有两个选择，要么删除，要么取，即2 (i – 1) 。
因此， K个连接的总数由下式给出：
ans = ans * (2(K*N) -1) / (2N – 1)
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
typedef long long LL;
const int MOD = 1000000007;
// Function to find the value of a^b
// modulo 1000000007
int exp_mod(LL a, LL b)
{
// Stores the resultant value a^b
LL ret = 1;
// Find the value of a^b
for (; b; b >>= 1, a = a * a % MOD) {
if (b & 1)
ret = ret * a % MOD;
}
return ret;
}
// Function to count the number of ways
// such that the formed number is divisible
// by 5 by removing digits
int countOfWays(string s, int k)
{
int N = s.size();
// Stores the count of ways
LL ans = 0;
// Find the count for string S
for (int i = 0; i < N; i++) {
// If the digit is 5 or 0
if (s[i] == '5' || s[i] == '0') {
ans = (ans + exp_mod(2, i)) % MOD;
}
}
// Find the count of string for K
// concatenation of string S
LL q = exp_mod(2, N);
LL qk = exp_mod(q, k);
LL inv = exp_mod(q - 1, MOD - 2);
// Find the total count
ans = ans * (qk - 1) % MOD;
ans = ans * inv % MOD;
return ans;
}
// Driver Code
int main()
{
string S = "1256";
int K = 1;
cout << countOfWays(S, K);
return 0;
} Java
// Java program for the above approach
class GFG
{
static long MOD = 1000000007;
// Function to find the value of a^b
// modulo 1000000007
public static long exp_mod(long a, long b) {
// Stores the resultant value a^b
long ret = 1;
// Find the value of a^b
for (; b > 0; b >>= 1, a = a * a % MOD) {
if ((b & 1) > 0)
ret = ret * a % MOD;
}
return ret;
}
// Function to count the number of ways
// such that the formed number is divisible
// by 5 by removing digits
public static long countOfWays(String s, int k) {
int N = s.length();
// Stores the count of ways
long ans = 0;
// Find the count for string S
for (int i = 0; i < N; i++) {
// If the digit is 5 or 0
if (s.charAt(i) == '5' || s.charAt(0) == '0') {
ans = (ans + exp_mod(2, i)) % MOD;
}
}
// Find the count of string for K
// concatenation of string S
long q = exp_mod(2, N);
long qk = exp_mod(q, k);
long inv = exp_mod(q - 1, MOD - 2);
// Find the total count
ans = ans * (qk - 1) % MOD;
ans = ans * inv % MOD;
return ans;
}
// Driver Code
public static void main(String args[]) {
String S = "1256";
int K = 1;
System.out.println(countOfWays(S, K));
}
}
// This code is contributed by _saurabh_jaiswal.Python3
# Python program for the above approach
MOD = 1000000007
# Function to find the value of a^b
# modulo 1000000007
def exp_mod(a, b):
# Stores the resultant value a^b
ret = 1
# Find the value of a^b
while(b):
if (b & 1):
ret = ret * a % MOD
b >>= 1
a = a * a % MOD
return ret
# Function to count the number of ways
# such that the formed number is divisible
# by 5 by removing digits
def countOfWays(s, k):
N = len(s)
# Stores the count of ways
ans = 0
# Find the count for string S
for i in range(N):
# If the digit is 5 or 0
if (s[i] == '5' or s[i] == '0'):
ans = (ans + exp_mod(2, i)) % MOD
# Find the count of string for K
# concatenation of string S
q = exp_mod(2, N)
qk = exp_mod(q, k)
inv = exp_mod(q - 1, MOD - 2)
# Find the total count
ans = ans * (qk - 1) % MOD
ans = ans * inv % MOD
return ans
# Driver Code
S = "1256"
K = 1
print(countOfWays(S, K))
# This code is contributed by shivaniC#
// C# program for the above approach
using System;
public class GFG
{
static long MOD = 1000000007;
// Function to find the value of a^b
// modulo 1000000007
public static long exp_mod(long a, long b) {
// Stores the resultant value a^b
long ret = 1;
// Find the value of a^b
for (; b > 0; b >>= 1, a = a * a % MOD) {
if ((b & 1) > 0)
ret = ret * a % MOD;
}
return ret;
}
// Function to count the number of ways
// such that the formed number is divisible
// by 5 by removing digits
public static long countOfWays(String s, int k) {
int N = s.Length;
// Stores the count of ways
long ans = 0;
// Find the count for string S
for (int i = 0; i < N; i++) {
// If the digit is 5 or 0
if (s[i] == '5' || s[0] == '0') {
ans = (ans + exp_mod(2, i)) % MOD;
}
}
// Find the count of string for K
// concatenation of string S
long q = exp_mod(2, N);
long qk = exp_mod(q, k);
long inv = exp_mod(q - 1, MOD - 2);
// Find the total count
ans = ans * (qk - 1) % MOD;
ans = ans * inv % MOD;
return ans;
}
// Driver Code
public static void Main(String []args) {
String S = "1256";
int K = 1;
Console.WriteLine(countOfWays(S, K));
}
}
// This code is contributed by shikhasingrajput输出：
4时间复杂度： O(N*log K)
辅助空间： O(1)