给定范围内任意对的最小乘积模 N
给定三个整数 L 、 R 和 N ，任务是求 (i * j) % N 的最小可能值，其中 L ≤ i < j ≤ R 。
示例:
输入: L = 2020，R = 2040，N = 2019 T3】输出:2 T6】说明: (2020 * 2021) % 2019 = 2
输入: L = 15，R = 30，N = 15 输出: 0 说明:如果该对元素中有一个是 15，那么所有这样的对的乘积将被 15 整除。因此，余数将为 0，这是最小可能值。
方法:找到 L 和 R 的区别就可以解决给定的问题。如果相差至少 N ，那么结果就是 0 。否则，迭代范围【L，R】，求最小乘积。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
#define ll long long
using namespace std;
// Function to return the minimum
// possible value of (i * j) % N
void minModulo(int L, int R, int N)
{
if (R - L < N) {
// Stores the minimum remainder
int ans = INT_MAX;
// Iterate from L to R
for (ll i = L; i <= R; i++)
// Iterate from L to R
for (ll j = L; j <= R; j++)
if (i != j)
ans = min(0ll + ans,
(i * j) % N);
// Print the minimum value
// of remainder
cout << ans;
}
// If R - L >= N
else {
cout << 0;
}
}
// Driver Code
int main()
{
int L = 6, R = 10, N = 2019;
minModulo(L, R, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG
{
// Function to return the minimum
// possible value of (i * j) % N
static void minModulo(int L, int R, int N)
{
if (R - L < N)
{
// Stores the minimum remainder
int ans = Integer.MAX_VALUE;
// Iterate from L to R
for (int i = L; i <= R; i++)
// Iterate from L to R
for (int j = L; j <= R; j++)
if (i != j)
ans = Math.min(ans, (i * j) % N);
// Print the minimum value
// of remainder
System.out.println(ans);
}
// If R - L >= N
else {
System.out.println(0);
}
}
// Driver Code
public static void main(String[] args)
{
int L = 6, R = 10, N = 2019;
minModulo(L, R, N);
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function to return the minimum
# possible value of (i * j) % N
def minModulo(L, R, N):
if (R - L < N):
# Stores the minimum remainder
ans = 10**9
# Iterate from L to R
for i in range(L, R + 1):
# Iterate from L to R
for j in range(L, R + 1):
if (i != j):
ans = min(ans, (i * j) % N)
# Print the minimum value
# of remainder
print (ans)
# If R - L >= N
else:
print (0)
# Driver Code
if __name__ == '__main__':
L, R, N = 6, 10, 2019
minModulo(L, R, N)
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to return the minimum
// possible value of (i * j) % N
static void minModulo(int L, int R, int N)
{
if (R - L < N)
{
// Stores the minimum remainder
int ans = Int32.MaxValue;
// Iterate from L to R
for(int i = L; i <= R; i++)
// Iterate from L to R
for(int j = L; j <= R; j++)
if (i != j)
ans = Math.Min(ans, (i * j) % N);
// Print the minimum value
// of remainder
Console.WriteLine(ans);
}
// If R - L >= N
else
{
Console.WriteLine(0);
}
}
// Driver Code
public static void Main(string[] args)
{
int L = 6, R = 10, N = 2019;
minModulo(L, R, N);
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// Javascript implementation
// for the above approach
// Function to return the minimum
// possible value of (i * j) % N
function minModulo(L, R, N)
{
if (R - L < N)
{
// Stores the minimum remainder
let ans = Number.MAX_VALUE;
// Iterate from L to R
for (let i = L; i <= R; i++)
// Iterate from L to R
for (let j = L; j <= R; j++)
if (i != j)
ans = Math.min(ans, (i * j) % N);
// Print the minimum value
// of remainder
document.write(ans);
}
// If R - L >= N
else {
document.write(0);
}
}
// Driver Code
let L = 6, R = 10, N = 2019;
minModulo(L, R, N);
// This code is contributed by susmitakundugoaldanga.
</script>
Output:
42
时间复杂度:O((R–L)2) 辅助空间: O(1)
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