題目來源：https://leetcode.com/contest/weekly-contest-106/problems/3sum-with-multiplicity/

問題描述

923. 3Sum With Multiplicity

Given an integer array A, and an integer target, return the number of tuples i, j, k  such that i < j < k and A[i] + A[j] + A[k] == target

As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Note:

1. 3 <= A.length <= 3000
2. 0 <= A[i] <= 100
3. 0 <= target <= 300

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題意

給定一組數，求其中三個數a1+a2+a3=target的可能的組合方式，要求三個數的數值可以相同，但必須是升序

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思路

注意到陣列中的數的取值範圍(0~100)遠小於陣列長度(0~3000)，故首先用計數排序把原數列轉化為0~100中各個數出現的個數。然後分“a1<a2<a3”“a1<a2=a3”“a1=a2<a3”“a1=a2=a3”討論。

注意儘管1e9+7在int範圍內，但結果一次增加的數值有可能超過int，所以結果和計數排序的陣列都要用long long.

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程式碼

class Solution {
public:
	const int MOD = (int)(1e9 + 7);
	long long n;
    long long a[105];

	int threeSumMulti(vector<int>& A, int target) {
		n = 0;
		int i, j, k, len = A.size();
        for (i=0; i<len; i++)
        {
            a[A[i]]++;
        }
		for (i = 0; i < 105; i++)
		{
			for (j = i+1; j < 105; j++)
			{
				for (k = j+1; k < 105; k++)
				{
					if (a[i] > 0 && a[j] > 0 && a[k] > 0 && i+j+k == target)
					{
						n += a[i]*a[j]*a[k];
                        n %= MOD;
					}
				}
			}
		}
        for (i=0; i<105; i++)
        {
            for (j=i+1; j<105; j++)
            {
                if (a[j]>=2 && a[i]>0 && i+2*j==target)
                {
                    n += a[i]*a[j]*(a[j]-1)/2;
                    n %= MOD;
                }
            }
        }
        for (i=0; i<105; i++)
        {
            for (j=i+1; j<105; j++)
            {
                if (a[j]>0 && a[i]>=2 && 2*i+j==target)
                {
                    n += a[j]*a[i]*(a[i]-1)/2;
                    n %= MOD;
                }
            }
        }
        for (i=0; i<105; i++)
        {
            if (a[i] >=3 && 3*i == target)
            {
                n += a[i]*(a[i]-1)*(a[i]-2)/6;
                n %= MOD;
            }
        }
		return n;
	}
};