习题8-18(uva-1619)-优快云博客

本文链接：https://blog.youkuaiyun.com/seanbill/article/details/117385987

本文介绍了一段C++代码，用于解决求解数组中，给定输入元素下，和可能超过整型范围的最大子数组乘积问题。通过维护左右边界，找到包含最大乘积的子数组范围。

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#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 1e5 + 7;
int input[maxn], l[maxn], r[maxn], n;
//和可能超过int
long long int sum[maxn] = { 0 };

int main()
{
	int kCase = 0;
	while (cin >> n && n) {
		for (int i = 1; i <= n; i++) {
			cin >> input[i];
			l[i] = r[i] = i; 
			sum[i] = sum[i - 1] + input[i];
		}
		
		input[0] = input[n + 1] = -1;
		for (int i = 1; i <= n; i++) {
			while (input[i] <= input[l[i] - 1])l[i] = l[l[i] - 1];
		}
		for (int i = n; i >= 1; i--) {
			while (input[i] <= input[r[i] + 1])r[i] = r[r[i] + 1];
		}
		long long int  ans = 0, ansL = 1, ansR = 1;
		for (int i = 1; i <= n; i++) {
			long long int  s = (long long int)input[i] * (sum[r[i]] - sum[l[i] - 1]);
			if (s  > ans || (s == ans &&  (r[i] - l[i]) < (ansR - ansL))) {
				ans = s;
				ansL = l[i];
				ansR = r[i];
			}
		}
		if (kCase++) cout << endl;
		cout << ans << endl << ansL << " " << ansR << endl;
	}
	return 0;
}