唉，TLE，提供一下方法吧

题目描述
有n个人，每个人都拥有一个喜悦值，接下来，任意三个人之间会进行一次交流，任意三个人交流所获得的贡献为三个人的喜悦值的和，求任意三个人交流所获得的贡献之和
输入格式：
第一行一个整数n
接下来n行，每行一个整数a，表示第i个人的喜悦值
输出格式：
按题目描述输出
样例输入1：
3
2
3
4
样例输出1：
9
约定：
n<=1000
1<=a<=10^100

#include <bits/stdc++.h>
class BigInt
{
public:
    BigInt(unsigned int n = 0)
    {
        if (n == 0)
            _data.push_back('0');
        else
        {
            std::string str;
            while (n)
            {
                str.push_back(n % 10 + '0');
                n /= 10;
            }
            _data.assign(str.rbegin(), str.rend());
        }
    }
  
    BigInt(const std::string &s) : _data(s) {}
  
private:
    std::string _data;
  
    friend BigInt operator+(const BigInt &lsh, const BigInt &rhs);
    friend std::istream &operator>>(std::istream &is, BigInt &a);
    friend std::ostream &operator<<(std::ostream &os, const BigInt &a);
};
  
BigInt operator+(const BigInt &lhs, const BigInt &rhs)
{
    std::string r;
    int carry = 0;
    std::string::const_reverse_iterator itr1 = lhs._data.rbegin();
    std::string::const_reverse_iterator itr2 = rhs._data.rbegin();
    while (itr1 != lhs._data.rend() && itr2 != rhs._data.rend())
    {
        int a = *itr1 - '0';
        int b = *itr2 - '0';
        int c = a + b + carry;
        if (c >= 10)
        {
            carry = c / 10;
            c %= 10;
        }
        else
        {
            carry = 0;
        }
        r.push_back(c + '0');
        ++itr1;
        ++itr2;
    }
    while (itr1 != lhs._data.rend())
    {
        int a = *itr1 - '0' + carry;
        if (a >= 10)
        {
            carry = a / 10;
            a %= 10;
        }
        else
        {
            carry = 0;
        }
        r.push_back(a + '0');
        ++itr1;
    }
    while (itr2 != rhs._data.rend())
    {
        int a = *itr2 - '0' + carry;
        if (a >= 10)
        {
            carry = a / 10;
            a %= 10;
        }
        else
        {
            carry = 0;
        }
        r.push_back(a + '0');
        ++itr2;
    }
    if (carry > 0)
        r.push_back(carry + '0');
    return BigInt(std::string(r.rbegin(), r.rend()));
}
  
std::istream &operator>>(std::istream &is, BigInt &a)
{
    is >> a._data;
    return is;
}
  
std::ostream &operator<<(std::ostream &os, const BigInt &a)
{
    os << a._data;
    return os;
}
  
int main()
{
    int n;
    std::cin >> n;
    std::vector<BigInt> a(n);
    for (int i = 0; i < n; i++)
        std::cin >> a[i];
    BigInt sum = 0;
    for (int i = 0; i < n - 2; i++)
        for (int j = i + 1; j < n - 1; j++)
            for (int k = j + 1; k < n; k++)
                sum = sum + a[i] + a[j] + a[k];
    std::cout << sum;
    return 0;
}//源自—GX—

想不出来如何优化

#include <iostream>
#include <string>
#include <iterator>
#include <vector>

class BigInt
{
public:
    BigInt(unsigned int n = 0)
    {
        if (n == 0)
            _data.push_back('0');
        else
        {
            while (n)
            {
                _data.push_back(n % 10 + '0');
                n /= 10;
            }
        }
    }

    BigInt(const std::string &s) : _data(s.rbegin(), s.rend()) {}

    BigInt &operator+=(const BigInt &other);
    BigInt &operator*=(const BigInt &other);

    BigInt operator+(const BigInt &other) const
    {
        BigInt t = *this;
        t += other;
        return t;
    }

    BigInt operator*(const BigInt &other) const
    {
        BigInt t = *this;
        t *= other;
        return t;
    }

private:
    std::string _data;

    friend std::istream &operator>>(std::istream &is, BigInt &a);
    friend std::ostream &operator<<(std::ostream &os, const BigInt &a);
};

BigInt &BigInt::operator+=(const BigInt &other)
{
    int carry = 0;
    std::string::iterator itr1 = _data.begin();
    std::string::const_iterator itr2 = other._data.begin();
    while (itr1 != _data.end() && itr2 != other._data.end())
    {
        int a = *itr1 - '0';
        int b = *itr2 - '0';
        int c = a + b + carry;
        carry = c / 10;
        c %= 10;
        *itr1 = c + '0';
        ++itr1;
        ++itr2;
    }
    while (itr1 != _data.end())
    {
        int a = *itr1 - '0' + carry;
        carry = a / 10;
        a %= 10;
        *itr1 = a + '0';
        ++itr1;
    }
    while (itr2 != other._data.end())
    {
        int a = *itr2 - '0' + carry;
        carry = a / 10;
        a %= 10;
        _data.push_back(a + '0');
        ++itr2;
    }
    if (carry > 0)
        _data.push_back(carry + '0');
    return *this;
}

BigInt &BigInt::operator*=(const BigInt &other)
{
    BigInt r;
    for (std::size_t i = 0; i < other._data.size(); i++)
    {
        int x = other._data[i] - '0';
        int carry = 0;
        BigInt t = *this;
        for (std::string::iterator itr = t._data.begin(); itr != t._data.end(); ++itr)
        {
            int y = *itr - '0';
            int a = x * y + carry;
            *itr = a % 10 + '0';
            carry = a / 10;
        }
        if (carry > 0)
            t._data.push_back(carry + '0');
        if (i > 0)
        {
            std::string zeros(i, '0');
            t._data.insert(t._data.begin(), zeros.begin(), zeros.end());
        }
        r += t;
    }
    *this = r;
    return *this;
}

std::istream &operator>>(std::istream &is, BigInt &a)
{
    std::string s;
    is >> s;
    a = s;
    return is;
}

std::ostream &operator<<(std::ostream &os, const BigInt &a)
{
    std::copy(a._data.rbegin(), a._data.rend(), std::ostream_iterator<char>(os));
    return os;
}

int main()
{
    int n;
    std::cin >> n;
    std::vector<BigInt> a(n);
    for (int i = 0; i < n; i++)
        std::cin >> a[i];
    BigInt sum = 0;
    BigInt num = (n - 1) * (n - 2) / 2;
    for (int i = 0; i < n; i++)
        sum += a[i] * num;
    std::cout << sum;
    return 0;
}

下面几个优化点，你可以考虑：

1. BigInt里的_data采用逆序存储数字（即个位在前，高位数在后），这样计算加法时就没必要逆序计算再把结果逆序过来。
2. 实现BigInt::operator+=()运算，直接在BigInt类_data上进行加法操作，省去了字符串拷贝，这样sum = sum + a[i] + a[j] + a[k];可以改为sum+=a[i]; sum+=a[j]; sum+=a[k];

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