给定两个整数X和Y ，找出数字之和为X且严格大于Y的最小数。
例子：
Input: X = 18, Y = 99
Output: 189
Explanation:
189 is the smallest number greater than 99 having sum of digits = 18.
Input: X = 12, Y = 72
Output: 75
Explanation:
75 is the smallest number greater than 72 that has sum of digits = 12.
朴素的方法：朴素的方法是从Y + 1迭代并检查是否有任何数字之和为X的数字。如果我们找到任何这样的数字，则打印该数字。
时间复杂度： O((R – Y)*log 10 N)，其中 R 是迭代之前的最大数，N 是[Y, R]范围内的数
辅助空间： O(1)
有效的方法：思想是从右到左遍历Y的数字，并尝试增加当前数字并向右更改数字，以使数字之和等于X 。以下是步骤：
- 如果我们考虑从右数第(k + 1)个数字并增加它，那么可以使k 个最低有效数字之和为[0, 9k]范围内的任何数字。
- 当找到这样的位置时，停止该过程并打印该迭代的数字。
- 如果k 个最低有效数字的总和为M （其中 0 ≤ M ≤ 9k），则贪婪地获得答案：
- 从右到左遍历并插入 9 并从数字总和中减去 9。
- 一次，总和小于 9，放置剩余的总和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return the minimum string
// of length d having the sum of digits s
string helper(int d, int s)
{
// Return a string of length d
string ans(d, '0');
for (int i = d - 1; i >= 0; i--) {
// Greedily put 9's in the end
if (s >= 9) {
ans[i] = '9';
s -= 9;
}
// Put remaining sum
else {
char c = (char)s + '0';
ans[i] = c;
s = 0;
}
}
return ans;
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
string findMin(int x, int Y)
{
// Convert number y to string
string y = to_string(Y);
int n = y.size();
vector p(n);
// Maintain prefix sum of digits
for (int i = 0; i < n; i++) {
p[i] = y[i] - '0';
if (i > 0)
p[i] += p[i - 1];
}
// Iterate over Y from the back where
// k is current length of suffix
for (int i = n - 1, k = 0;; i--, k++) {
// Stores current digit
int d = 0;
if (i >= 0)
d = y[i] - '0';
// Increase current digit
for (int j = d + 1; j <= 9; j++) {
// Sum upto current prefix
int r = (i > 0) * p[i - 1] + j;
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 and x - r <= 9 * k) {
// Find suffix of length k
// having sum of digits x-r
string suf = helper(k, x - r);
string pre = "";
if (i > 0)
pre = y.substr(0, i);
// Append current character
char cur = (char)j + '0';
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver Code
int main()
{
// Given Number and Sum
int x = 18;
int y = 99;
// Function Call
cout << findMin(x, y) << endl;
return 0;
} Java
// Java program for the above approach
import java.util.*;
@SuppressWarnings("unchecked")
class GFG{
// Function to return the minimum String
// of length d having the sum of digits s
static String helper(int d, int s)
{
// Return a String of length d
StringBuilder ans = new StringBuilder();
for(int i = 0; i < d; i++)
{
ans.append("0");
}
for(int i = d - 1; i >= 0; i--)
{
// Greedily put 9's in the end
if (s >= 9)
{
ans.setCharAt(i,'9');
s -= 9;
}
// Put remaining sum
else
{
char c = (char)(s + (int)'0');
ans.setCharAt(i, c);
s = 0;
}
}
return ans.toString();
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
static String findMin(int x, int Y)
{
// Convert number y to String
String y = Integer.toString(Y);
int n = y.length();
ArrayList p = new ArrayList();
for(int i = 0; i < n; i++)
{
p.add(0);
}
// Maintain prefix sum of digits
for(int i = 0; i < n; i++)
{
p.add(i, (int)((int) y.charAt(i) - (int)'0'));
if (i > 0)
{
p.add(i, (int)p.get(i) +
(int)p.get(i - 1));
}
}
// Iterate over Y from the back where
// k is current length of suffix
for(int i = n - 1, k = 0;; i--, k++)
{
// Stores current digit
int d = 0;
if (i >= 0)
{
d = (int) y.charAt(i) - (int)'0';
}
// Increase current digit
for(int j = d + 1; j <= 9; j++)
{
int r = j;
// Sum upto current prefix
if (i > 0)
{
r += (int) p.get(i - 1);
}
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 && x - r <= 9 * k)
{
// Find suffix of length k
// having sum of digits x-r
String suf = helper(k, x - r);
String pre = "";
if (i > 0)
pre = y.substring(0, i);
// Append current character
char cur = (char)(j + (int)'0');
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver code
public static void main(String[] arg)
{
// Given number and sum
int x = 18;
int y = 99;
// Function call
System.out.print(findMin(x, y));
}
}
// This code is contributed by pratham76Python3
# Python3 program for the
# above approach
# Function to return the
# minimum string of length
# d having the sum of digits s
def helper(d, s):
# Return a string of
# length d
ans = ['0'] * d
for i in range(d - 1,
-1, -1):
# Greedily put 9's
# in the end
if (s >= 9):
ans[i] = '9'
s -= 9
# Put remaining sum
else:
c = chr(s +
ord('0'))
ans[i] = c;
s = 0;
return ''.join(ans);
# Function to find the
# smallest number greater
# than Y whose sum of
# digits is X
def findMin(x, Y):
# Convert number y
# to string
y = str(Y);
n = len(y)
p = [0] * n
# Maintain prefix sum
# of digits
for i in range(n):
p[i] = (ord(y[i]) -
ord('0'))
if (i > 0):
p[i] += p[i - 1];
# Iterate over Y from the
# back where k is current
# length of suffix
n - 1
k = 0
while True:
# Stores current digit
d = 0;
if (i >= 0):
d = (ord(y[i]) -
ord('0'))
# Increase current
# digit
for j in range(d + 1,
10):
# Sum upto current
# prefix
r = ((i > 0) *
p[i - 1] + j);
# Return answer if
# remaining sum can
# be obtained in suffix
if (x - r >= 0 and
x - r <= 9 * k):
# Find suffix of length
# k having sum of digits
# x-r
suf = helper(k,
x - r);
pre = "";
if (i > 0):
pre = y[0 : i]
# Append current
# character
cur = chr(j +
ord('0'))
pre += cur;
# Return the result
return pre + suf;
i -= 1
k += 1
# Driver Code
if __name__ == "__main__":
# Given Number and Sum
x = 18;
y = 99;
# Function Call
print ( findMin(x, y))
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Text;
using System.Collections;
class GFG{
// Function to return the minimum string
// of length d having the sum of digits s
static string helper(int d, int s)
{
// Return a string of length d
StringBuilder ans = new StringBuilder();
for(int i = 0; i < d; i++)
{
ans.Append("0");
}
for(int i = d - 1; i >= 0; i--)
{
// Greedily put 9's in the end
if (s >= 9)
{
ans[i] = '9';
s -= 9;
}
// Put remaining sum
else
{
char c = (char)(s + (int)'0');
ans[i] = c;
s = 0;
}
}
return ans.ToString();
}
// Function to find the smallest
// number greater than Y
// whose sum of digits is X
static string findMin(int x, int Y)
{
// Convert number y to string
string y = Y.ToString();
int n = y.Length;
ArrayList p = new ArrayList();
for(int i = 0; i < n; i++)
{
p.Add(0);
}
// Maintain prefix sum of digits
for(int i = 0; i < n; i++)
{
p[i] = (int)((int) y[i] - (int)'0');
if (i > 0)
{
p[i] = (int)p[i] +
(int)p[i - 1];
}
}
// Iterate over Y from the back where
// k is current length of suffix
for(int i = n - 1, k = 0;; i--, k++)
{
// Stores current digit
int d = 0;
if (i >= 0)
{
d = (int) y[i] - (int)'0';
}
// Increase current digit
for(int j = d + 1; j <= 9; j++)
{
int r = j;
// Sum upto current prefix
if (i > 0)
{
r += (int) p[i - 1];
}
// Return answer if remaining
// sum can be obtained in suffix
if (x - r >= 0 && x - r <= 9 * k)
{
// Find suffix of length k
// having sum of digits x-r
string suf = helper(k, x - r);
string pre = "";
if (i > 0)
pre = y.Substring(0, i);
// Append current character
char cur = (char)(j + (int)'0');
pre += cur;
// Return the result
return pre + suf;
}
}
}
}
// Driver code
public static void Main(string[] arg)
{
// Given number and sum
int x = 18;
int y = 99;
// Function call
Console.Write(findMin(x, y));
}
}
// This code is contributed by rutvik_56Javascript
输出：
189时间复杂度： O(log 10 Y)
辅助空间： O(log 10 Y)