uva 10747 - Maximum Subsequence(贪心)
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题目链接:10747 - Maximum Subsequence
题目大意:给出n个数,从中选取k个,使得乘积最大,并且尽量使和最大。
解题思路:首先按照数的绝对值大小排序。然后就要分三大类情况讨论:
(1)前k个中选到0:如果选到0的话,乘积一定是0,所以尽量选大的数,让和变大。
(2)前k个中选到负数的个数为偶数:这样的话直接输出答案(一定为最优解)
(3)前k个中选到的负数个数为奇数:这类情况比较复杂,还要分成两个子类:
a)k个中没有正数(意味着不能用未选的数中的负数来替换):
如果还正数:优先用正数替换最小的负数;否则注定乘积为负数或者0:选最大的k个。
b)k个中有正数:
还有正数和负数:比较最优; 只有正数:选用负数; 只有负数:选用正数; 只有0:选最大的k个,注意0一定要选。
对应的数据:
6 3 -5 -4 -3 -2 2 1
6 3 -5 -4 -3 -2 -1 0
6 3 -5 -4 -3 -2 -2 -1
6 3 5 4 -3 2 -1 1
6 3 5 -4 3 -2 1 1
6 3 5 4 -4 -2 2 1
6 3 5 4 -3 1 1 1
6 3 5 4 -3 -1 -1 -1
4 3 5 4 -3 0
答案:
-7
-3
-5
11
-1
11
10
1
9
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <algorithm>using namespace std;const int N = 10005;int n, k, num[N];bool cmp(const int& a, const int& b) {int p = abs(a), q = abs(b);if (p != q) return p > q;else return a > b;}void init() {for (int i = 0; i < n; i++) scanf("%d", &num[i]);sort(num, num + n, cmp);}int handle(int t) {sort(num, num + n);int sum = 0;for (int i = 1; i <= t; i++) {sum += num[n-i];}return sum;}int solve() {int sum = 0, cnt = 0, p = 0, q = 0;bool flag = false;for (int i = 0; i < k; i++) {if (num[i] == 0) {flag = true; break;} else if (num[i] < 0) {cnt++; q = num[i];} else p = num[i];sum += num[i];}if (flag) return handle(k);else if (cnt % 2 && k < n) {int x = 0, y = 0, bo = 0;for (int i = k; i < n; i++) if (num[i] > 0) {x = num[i]; break;}for (int i = k; i < n; i++) if (num[i] < 0) {y = num[i]; break;}for (int i = k; i < n; i++) if (num[i] == 0) {bo = 1; break;}if (p == 0) {if (x) sum = sum - q + x;else sum = handle(k);} else {if (x == 0 && y == 0) sum = handle(k - 1);else if (x == 0) sum = sum - p + y;else if (y == 0) sum = sum - q + x;else if (x * p >= y * q) sum = sum - q + x;else sum = sum - p + y;}}return sum;}int main() {while (scanf("%d%d", &n, &k) == 2 && n + k) {init();printf("%d\n", solve());}return 0;}
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