HackerRank - "Maximise Sum"

最新推荐文章于 2024-06-16 17:06:51 发布

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最新推荐文章于 2024-06-16 17:06:51 发布

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原文链接：http://www.cnblogs.com/tonix/p/4522240.html

本文探讨了模运算版本的连续子数组最大和问题，并介绍了关键技巧，如模运算的滚动特性、`std::lower_bound()`在`std::set`中的应用。文章通过实例说明了如何在模运算约束下求解最大连续子数组和，提供了快速正确的解决方案。

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It is modulo version of "max sum of contiguous subarray". So similar pattern can be applied too, but with tricks regarding to modulo ops.

I checked editorials, and found that, the quick\correct solution requires sharp insight into internal characteristics and mechanism of one problem... which I'm still lack with...

Modulo, you should feel of it is as a roll-over game. If within the scope, like (3 - 1)%3, there's nothing special; and if beyond the scope like (1 - 3)%3, you get it roll-over from the beginning..

And another key trick is, std::lower_bound() upon std::set is slower than set::lower_bound()..

#include <cmath>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <string>
#include <climits>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
using namespace std;

int dp(vector<int> &in, int m)
{
    int ret = 0;
    size_t len = in.size();

    vector<int> dp(len, 0);

    for (int ilen = 1; ilen <= len; ilen ++)
    for (int i = 0; i <= len - ilen; i++)
    {
        dp[i] += in[i + ilen - 1];
        ret = std::max(ret, dp[i] % m);
    }
    return ret;
}

int main()
{
    int t; cin >> t;
    while (t--)
    {
        long long n, m; cin >> n >> m;
    
        long long x, prefix = 0, ret = 0;
        set<long long> s;
        s.insert(0);
        for (int i = 0; i < n; i++)
        {
            cin >> x;
            prefix = (prefix + x) % m;
            
            //    Case 1: prefix - 0 
            //        (for smaller prefixes)
            ret = std::max(ret, prefix); 

            //    Case 2: prefix - smallest_larer + m 
            //        (for larger prefixes)
            auto it = s.lower_bound(prefix + 1);
            if (it != s.end())
            {
                ret = std::max(ret, prefix - *it + m);
            }

            s.insert(prefix);
        }
        cout << ret << endl;
    }
    return 0;
}