Jumping Cows

最新推荐文章于 2020-04-10 20:45:42 发布

转载最新推荐文章于 2020-04-10 20:45:42 发布 · 238 阅读

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原文链接：http://www.cnblogs.com/lavender913/p/3310851.html

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#java

Jumping Cows

时间限制(普通/Java):3000MS/10000MS 运行内存限制:65536KByte

描述

Farmer John's cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump.
The local witch doctor has mixed up P (1 <= P <= 150,000) potions to aid the cows in their quest to jump. These potions must be administered exactly in the order they were created, though some may be skipped.
Each potion has a 'strength' (1 <= strength <= 500) that enhances the cows' jumping ability. Taking a potion during an odd time step increases the cows' jump; taking a potion during an even time step decreases the jump. Before taking any potions the cows' jumping ability is, of course, 0.
No potion can be taken twice, and once the cow has begun taking potions, one potion must be taken during each time step, starting at time 1. One or more potions may be skipped in each turn.
Determine which potions to take to get the highest jump.

输入

* Line 1: A single integer, P
* Lines 2..P+1: Each line contains a single integer that is the strength of a potion. Line 2 gives the strength of the first potion; line 3 gives the strength of the second potion; and so on.

输出

* Line 1: A single integer that is the maximum possible jump.

样例输入

样例输出

题目大意：题目要求从输入的p个数中任意选出n(n<=p)个数按输入顺序排好，假设给n个数1至N编号，若i(1<=i<=N)是偶数，则将第i个数变为负数，最后求n个数的最大和。

#include<iostream>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;

int main() {
    int n,i,j,x;
    int ans,sum;
    while(~scanf("%d",&n)) {
        scanf("%d",&x);
        ans = x;
        sum = 0;
        for(i=1; i<n; i++) {
            scanf("%d",&x);
            if(sum+x>ans)
                ans = sum+x;
            if(ans-x>sum)
                sum = ans-x;
        }
        if(ans>sum)
            printf("%d\n",ans);
        else
            printf("%d\n",sum);
    }
    return 0;
}