给定一个数字(如字符串)和两个整数a和b，请将字符串分成两个非空部分，以使第一部分可被a整除，第二部分可被b整除。如果字符串不能分为两个非空部分，则输出“ NO”，否则将两个部分打印为“ YES”。
例子：
Input : str = "123", a = 12, b = 3
Output : YES
12, 3
"12" is divisible by a and "3" is
divisible by b.
Input : str = "1200", a = 4, b = 3
Output : YES
12, 00
Input : str = "125", a = 12, b = 3
Output : NO
一个简单的解决方案是在所有点周围一个一个的分区阵列。对于每个分区，检查其左和右是否分别可被a和b整除。如果是，请打印左右部分，然后返回。
一种有效的解决方案是进行一些预处理，并通过从左到右扫描字符串，并通过“ b”从右到左扫描模，以“ a”保存模除。
如果我们知道从0到i的前缀的余数，除以a，则可以使用以下公式计算从0到i + 1的前缀的余数。
lr [i + 1] =(lr [i] * 10 + str [i] -‘0’)％a。
同样，可以通过从右到左扫描找到以b为模的模。我们创建另一个rl []来存储从右到左带有b的余数。
一旦我们预先计算了两个余数，就可以轻松地将分割字符串分为两部分。
C++
// C++ program to check if a string can be splitted
// into two strings such that one is divisible by 'a'
// and other is divisible by 'b'.
#include
using namespace std;
// Finds if it is possible to partition str
// into two parts such that first part is
// divisible by a and second part is divisible
// by b.
void findDivision(string &str, int a, int b)
{
int len = str.length();
// Create an array of size len+1 and initialize
// it with 0.
// Store remainders from left to right when
// divided by 'a'
vector lr(len+1, 0);
lr[0] = (str[0] - '0')%a;
for (int i=1; i rl(len+1, 0);
rl[len-1] = (str[len-1] - '0')%b;
int power10 = 10;
for (int i= len-2; i>=0; i--)
{
rl[i] = (rl[i+1] + (str[i]-'0')*power10)%b;
power10 = (power10 * 10) % b;
}
// Find a point that can partition a number
for (int i=0; i Java
// Java program to check if a string can be splitted
// into two strings such that one is divisible by 'a'
// and other is divisible by 'b'.
class GFG
{
// Finds if it is possible to partition str
// into two parts such that first part is
// divisible by a and second part is divisible
// by b.
static void findDivision(String str, int a, int b)
{
int len = str.length();
// Create an array of size len+1 and initialize
// it with 0.
// Store remainders from left to right when
// divided by 'a'
int[] lr = new int[len + 1];
lr[0] = ((int)str.charAt(0) - (int)'0')%a;
for (int i = 1; i < len; i++)
lr[i] = ((lr[i - 1] * 10) % a +
((int)str.charAt(i)-(int)'0')) % a;
// Compute remainders from right to left when
// divided by 'b'
int[] rl = new int[len + 1];
rl[len - 1] = ((int)str.charAt(len - 1) -
(int)'0') % b;
int power10 = 10;
for (int i= len - 2; i >= 0; i--)
{
rl[i] = (rl[i + 1] + ((int)str.charAt(i) -
(int)'0') * power10) % b;
power10 = (power10 * 10) % b;
}
// Find a point that can partition a number
for (int i = 0; i < len - 1; i++)
{
// If split is not possible at this point
if (lr[i] != 0)
continue;
// We can split at i if one of the following
// two is true.
// a) All charactes after str.charAt(i] are 0
// b) String after str.charAt(i] is divisible by b, i.e.,
// str.charAt(i+1..n-1] is divisible by b.
if (rl[i + 1] == 0)
{
System.out.println("YES");
for (int k = 0; k <= i; k++)
System.out.print(str.charAt(k));
System.out.print(", ");
for (int k = i + 1; k < len; k++)
System.out.print(str.charAt(k));
return;
}
}
System.out.println("NO");
}
// Driver code
public static void main (String[] args)
{
String str = "123";
int a = 12, b = 3;
findDivision(str, a, b);
}
}
// This code is contributed by mitsPython3
# Python3 program to check if a can be splitted
# into two strings such that one is divisible by 'a'
# and other is divisible by 'b'.
# Finds if it is possible to partition str
# into two parts such that first part is
# divisible by a and second part is divisible
# by b.
def findDivision(str, a, b):
lenn = len(str)
# Create an array of size lenn+1 and
# initialize it with 0.
# Store remainders from left to right
# when divided by 'a'
lr = [0] * (lenn + 1)
lr[0] = (int(str[0]))%a
for i in range(1, lenn):
lr[i] = ((lr[i - 1] * 10) % a + \
int(str[i])) % a
# Compute remainders from right to left
# when divided by 'b'
rl = [0] * (lenn + 1)
rl[lenn - 1] = int(str[lenn - 1]) % b
power10 = 10
for i in range(lenn - 2, -1, -1):
rl[i] = (rl[i + 1] + int(str[i]) * power10) % b
power10 = (power10 * 10) % b
# Find a pothat can partition a number
for i in range(0, lenn - 1):
# If split is not possible at this point
if (lr[i] != 0):
continue
# We can split at i if one of the following
# two is true.
# a) All charactes after str[i] are 0
# b) after str[i] is divisible by b, i.e.,
# str[i+1..n-1] is divisible by b.
if (rl[i + 1] == 0):
print("YES")
for k in range(0, i + 1):
print(str[k], end = "")
print(",", end = " ")
for i in range(i + 1, lenn):
print(str[k], end = "")
return
print("NO")
# Driver code
str = "123"
a, b = 12, 3
findDivision(str, a, b)
# This code is contributed by SHUBHAMSINGH10C#
// C# program to check if a string can be splitted
// into two strings such that one is divisible by 'a'
// and other is divisible by 'b'.
using System;
class GFG
{
// Finds if it is possible to partition str
// into two parts such that first part is
// divisible by a and second part is divisible
// by b.
static void findDivision(string str, int a, int b)
{
int len = str.Length;
// Create an array of size len+1 and initialize
// it with 0.
// Store remainders from left to right when
// divided by 'a'
int[] lr = new int[len + 1];
lr[0] = ((int)str[0] - (int)'0')%a;
for (int i = 1; i < len; i++)
lr[i] = ((lr[i - 1] * 10) % a +
((int)str[i] - (int)'0')) % a;
// Compute remainders from right to left when
// divided by 'b'
int[] rl = new int[len + 1];
rl[len - 1] = ((int)str[len - 1] - (int)'0') % b;
int power10 = 10;
for (int i= len - 2; i >= 0; i--)
{
rl[i] = (rl[i + 1] + ((int)str[i] -
(int)'0') * power10) % b;
power10 = (power10 * 10) % b;
}
// Find a point that can partition a number
for (int i = 0; i < len - 1; i++)
{
// If split is not possible at this point
if (lr[i] != 0)
continue;
// We can split at i if one of the following
// two is true.
// a) All charactes after str[i] are 0
// b) String after str[i] is divisible by b, i.e.,
// str[i+1..n-1] is divisible by b.
if (rl[i + 1] == 0)
{
Console.WriteLine("YES");
for (int k = 0; k <= i; k++)
Console.Write(str[k]);
Console.Write(", ");
for (int k = i + 1; k < len; k++)
Console.Write(str[k]);
return;
}
}
Console.WriteLine("NO");
}
// Driver code
static void Main()
{
string str = "123";
int a = 12, b = 3;
findDivision(str, a, b);
}
}
// This code is contributed by mits输出 ：
YES
12, 3时间复杂度： O(len)其中len是输入数字字符串的长度。