给定整数N和整数S ,任务是找到大于或等于N的最小数字,以使其位数之和不超过S。
例子:
Input: N = 3, S = 2
Output: 10
Explanation: Sum of digits of 10 is 1, which is less than 2.
Input: N = 19, S = 3
Output: 20
Explanation: Sum of digits of 20 is 2, which is less than 3.
方法:可以使用贪婪方法解决问题。请按照以下步骤解决问题。
- 检查N的位数之和不超过S ,返回N。
- 初始化一个变量,比如说ans与1等于给定的整数N和k ,以存储10的幂。
- 整数范围内最多可以有10位数字。
- 从i = 0迭代到8 。在每次迭代时,将最后一位计算为(ans / k)%10 。
- 使最后一位数字为0的总和为k *((10-last_digit)%10) 。将其添加到ans 。
- 检查ans的位数之和。如果不超过S ,则打印ans并中断。否则,更新k作为K = K * 10,并重复上述步骤。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate sum
// digits of n
int sum(int n)
{
int res = 0;
while (n > 0) {
res += n % 10;
n /= 10;
}
return res;
}
// Function to find the smallest
// possible integer satisfying the
// given condition
int smallestNumber(int n, int s)
{
// If the sum of digits
// is already smaller than S
if (sum(n) <= s) {
return n;
}
// Initialize variables
int ans = n, k = 1;
for (int i = 0; i < 9; ++i) {
// Finding last kth digit
int digit = (ans / k) % 10;
// Add remaining to make digit 0
int add = k * ((10 - digit) % 10);
ans += add;
// If sum of digits
// does not exceed S
if (sum(ans) <= s) {
break;
}
// Update k
k *= 10;
}
return ans;
}
// Driver Code
int main()
{
// Given N and S
int N = 3, S = 2;
// Function call
cout << smallestNumber(N, S) << endl;
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to calculate sum
// digits of n
static int sum(int n)
{
int res = 0;
while (n > 0)
{
res += n % 10;
n /= 10;
}
return res;
}
// Function to find the smallest
// possible integer satisfying the
// given condition
static int smallestNumber(int n, int s)
{
// If the sum of digits
// is already smaller than S
if (sum(n) <= s)
{
return n;
}
// Initialize variables
int ans = n, k = 1;
for(int i = 0; i < 9; ++i)
{
// Finding last kth digit
int digit = (ans / k) % 10;
// Add remaining to make digit 0
int add = k * ((10 - digit) % 10);
ans += add;
// If sum of digits
// does not exceed S
if (sum(ans) <= s)
{
break;
}
// Update k
k *= 10;
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given N and S
int N = 3, S = 2;
// Function call
System.out.println(smallestNumber(N, S));
}
}
// This code is contributed by akhilsaini
Python3
# Python program for the above approach
# Function to calculate
# sum of digits of n
def sum(n):
sm = 0
while(n > 0):
sm += n % 10
n //= 10
return sm
# Function to find the smallest
# possible integer satisfying the
# given condition
def smallestNumber(n, s):
# If sum of digits is
# already smaller than s
if(sum(n) <= s):
return n
# Initialize variables
ans, k = n, 1
for i in range(9):
# Find the k-th digit
digit = (ans // k) % 10
# Add remaining
add = k * ((10 - digit) % 10)
ans += add
# If sum of digits
# does not exceed s
if(sum(ans) <= s):
break
# Update K
k *= 10
# Return answer
return ans
# Driver Code
# Given N and S
n, s = 3, 2
# Function call
print(smallestNumber(n, s))
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate sum
// digits of n
static int sum(int n)
{
int res = 0;
while (n > 0)
{
res += n % 10;
n /= 10;
}
return res;
}
// Function to find the smallest
// possible integer satisfying the
// given condition
static int smallestNumber(int n, int s)
{
// If the sum of digits
// is already smaller than S
if (sum(n) <= s)
{
return n;
}
// Initialize variables
int ans = n, k = 1;
for(int i = 0; i < 9; ++i)
{
// Finding last kth digit
int digit = (ans / k) % 10;
// Add remaining to make digit 0
int add = k * ((10 - digit) % 10);
ans += add;
// If sum of digits
// does not exceed S
if (sum(ans) <= s)
{
break;
}
// Update k
k *= 10;
}
return ans;
}
// Driver Code
public static void Main()
{
// Given N and S
int N = 3, S = 2;
// Function call
Console.WriteLine(smallestNumber(N, S));
}
}
// This code is contributed by akhilsaini
Javascript
输出:
10
时间复杂度: O(log 2 10 (N)),其中N是给定的整数。
空间复杂度: O(1)