给定n个数，找到一个最大区间使得区间内所有数的和能被n整除，用c++给出代码

以下是用C++实现的代码示例：

#include <iostream>
#include <vector>
#include <unordered_map>

std::pair<int, int> findMaxInterval(std::vector<int>& arr, int n) {
    std::vector<int> prefixSum(arr.size() + 1, 0);
    for (int i = 1; i <= arr.size(); i++) {
        prefixSum[i] = (prefixSum[i - 1] + arr[i - 1]) % n;
    }

    int maxLength = 0;
    int intervalStart = 0;
    int intervalEnd = 0;
    std::unordered_map<int, int> prefixSumHash;
    for (int i = 0; i < prefixSum.size(); i++) {
        if (prefixSum[i] == 0) {
            maxLength = i;
            intervalStart = 0;
            intervalEnd = i - 1;
        } else if (prefixSumHash.count(prefixSum[i])) {
            if (i - prefixSumHash[prefixSum[i]] > maxLength) {
                maxLength = i - prefixSumHash[prefixSum[i]];
                intervalStart = prefixSumHash[prefixSum[i]] + 1;
                intervalEnd = i - 1;
            }
        } else {
            prefixSumHash[prefixSum[i]] = i;
        }
    }

    return std::make_pair(intervalStart, intervalEnd);
}

int main() {
    std::vector<int> arr = {4, 3, 1, 6, 7};
    int n = 3;

    std::pair<int, int> result = findMaxInterval(arr, n);
    int start = result.first;
    int end = result.second;

    std::cout << "最大区间起始索引: " << start << std::endl;
    std::cout << "最大区间结束索引: " << end << std::endl;

    return 0;
}

这段代码也是基于前缀和的思想，使用了unordered_map来保存前缀和和其对应的索引。时间复杂度为O(n)，其中n是给定数组的长度。