求所有有效对的 a[i]%a[j]的和
原文:https://www . geesforgeks . org/find-sum-of-aiaj-for-all-valid-pairs/
给定一个大小为 N 的数组 arr[] 。任务是找到所有有效对的 arr[i] % arr[j] 之和。答案可以很大。于是，输出答案模 1000000007 例:
输入: arr[] = {1，2，3} 输出:5 (1% 1)+(1% 2)+(1% 3)+(2% 1)+(2% 2) +(2% 3)+(3% 1)+(3% 2)+(3% 3)= 5 输入: arr[] = {1，2，4，4，4} 输出:
方法:存储每个元素的频率，在频率数组上运行嵌套循环，找到需要的答案。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define mod (int)(1e9 + 7)
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
int Sum_Modulo(int a[], int n)
{
int max = *max_element(a, a + n);
// To store the frequency of each element
int cnt[max + 1] = { 0 };
// Store the frequency of each element
for (int i = 0; i < n; i++)
cnt[a[i]]++;
// To store the required answer
long long ans = 0;
// For all valid pairs
for (int i = 1; i <= max; i++) {
for (int j = 1; j <= max; j++) {
// Update the count
ans = ans + cnt[i] * cnt[j] * (i % j);
ans = ans % mod;
}
}
return (int)(ans);
}
// Driver code
int main()
{
int a[] = { 1, 2, 3 };
int n = sizeof(a) / sizeof(a[0]);
cout << Sum_Modulo(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
static int mod = (int)(1e9 + 7);
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
static int Sum_Modulo(int a[], int n)
{
int max = Arrays.stream(a).max().getAsInt();
// To store the frequency of each element
int []cnt=new int[max + 1];
// Store the frequency of each element
for (int i = 0; i < n; i++)
cnt[a[i]]++;
// To store the required answer
long ans = 0;
// For all valid pairs
for (int i = 1; i <= max; i++)
{
for (int j = 1; j <= max; j++)
{
// Update the count
ans = ans + cnt[i] *
cnt[j] * (i % j);
ans = ans % mod;
}
}
return (int)(ans);
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 2, 3 };
int n = a.length;
System.out.println(Sum_Modulo(a, n));
}
}
// This code is contributed
// by PrinciRaj1992
Python 3
# Python3 implementation of the approach
mod = 10**9 + 7
# Function to return the sum of
# (a[i] % a[j]) for all valid pairs
def Sum_Modulo(a, n):
Max = max(a)
# To store the frequency of each element
cnt = [0 for i in range(Max + 1)]
# Store the frequency of each element
for i in a:
cnt[i] += 1
# To store the required answer
ans = 0
# For all valid pairs
for i in range(1, Max + 1):
for j in range(1, Max + 1):
# Update the count
ans = ans + cnt[i] * \
cnt[j] * (i % j)
ans = ans % mod
return ans
# Driver code
a = [1, 2, 3]
n = len(a)
print(Sum_Modulo(a, n))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
static int mod = (int)(1e9 + 7);
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
static int Sum_Modulo(int []a, int n)
{
int max = a.Max();
// To store the frequency of each element
int []cnt = new int[max + 1];
// Store the frequency of each element
for (int i = 0; i < n; i++)
cnt[a[i]]++;
// To store the required answer
long ans = 0;
// For all valid pairs
for (int i = 1; i <= max; i++)
{
for (int j = 1; j <= max; j++)
{
// Update the count
ans = ans + cnt[i] *
cnt[j] * (i % j);
ans = ans % mod;
}
}
return (int)(ans);
}
// Driver code
public static void Main(String[] args)
{
int []a = { 1, 2, 3 };
int n = a.Length;
Console.WriteLine(Sum_Modulo(a, n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript implementation of the approach
let mod = 1e9 + 7
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
function Sum_Modulo(a, n)
{
let max = a.sort((a, b) => b - a)[0];
// To store the frequency of each element
let cnt = new Array(max + 1).fill(0);
// Store the frequency of each element
for (let i = 0; i < n; i++)
cnt[a[i]]++;
// To store the required answer
let ans = 0;
// For all valid pairs
for (let i = 1; i <= max; i++) {
for (let j = 1; j <= max; j++) {
// Update the count
ans = ans + cnt[i] * cnt[j] * (i % j);
ans = ans % mod;
}
}
return (ans);
}
// Driver code
let a = [1, 2, 3];
let n = a.length;
document.write(Sum_Modulo(a, n));
// This code is contributed by _saurabh_jaiswal
</script>
Output:
5
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