You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence.
InputThe first line contains integer number n (1 ≤ n ≤ 105).
The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum.
OutputPrint sum of resulting subseqeuence.
Examples input 4 -2 2 -3 1 output 3 input 3 2 -5 -3 output -1 NoteIn the first example sum of the second and the fourth elements is 3.
题意概述:给你n个数,要你它们子序列中求和为奇数的序列的最大值。 解题思路:如果数为偶数且大于0则肯定要选,奇数则讨论。对所有的奇数按最大值排序后求和,那么每加一个奇数就更新一下最大值。 AC代码: #include <bits/stdc++.h> #define INF 0x3f3f3f3f #define maxn 100100 #define lson root << 1 #define rson root << 1 | 1 #define lent (t[root].r - t[root].l + 1) #define lenl (t[lson].r - t[lson].l + 1) #define lenr (t[rson].r - t[rson].l + 1) #define N 1111 #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) using namespace std; const int mod = 1e9 + 7; typedef long long ll; typedef unsigned long long ull; vector<int> odd; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif int n; while (~scanf("%d", &n)) { odd.clear(); int ans = 0; for (int i = 0; i < n; ++i) { int x; scanf("%d", &x); if (x & 1) odd.push_back(x); else if (x > 0) ans += x; } sort(odd.begin(), odd.end(), greater<int>()); int sz = odd.size(); int sum = 0, res = -INF; for (int i = 0; i < sz; i++) { sum += odd[i]; if (sum & 1) res = max(res, sum + ans); } printf("%d\n", res); } #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; }