核心:从左往右遍历,找递增序列,从右往左遍历,找递增序列,输出两者最大值

本质:熊猫重的喝的奶要比别人至少多100ml

满分

#include <iostream>
#include <cstring>

using namespace std;

const int N = 10010;
int p[N], l[N], r[N];

int main()
{
    int m, num = 2, sum = 0;
    cin >> m;
    for(int i = 0; i < m; i++) cin >> p[i];
    l[0] = num;
    for(int i = 1; i < m; i++)
    {
        if (p[i] > p[i - 1]) l[i] = l[i - 1] + 1;
        else if (p[i] == p[i - 1]) l[i] = l[i - 1];
        else l[i] = 2;
    }
    num = 2;
    r[m - 1] = num;
    for (int i = m - 2; i >= 0; i--)
    {
        if (p[i] > p[i + 1]) r[i] = r[i + 1] + 1;
        else if (p[i] == p[i + 1]) r[i] = r[i + 1];
        else r[i] = 2;
    }

    for(int i = 0; i < m; i++)
    {
        sum += max(l[i], r[i]);
    }
    printf("%d", sum * 100);
}

17分代码 最后一个测试点错误

/**
题意: 胖胖的熊猫需要多的牛奶

做法:两次循环即可,先从左到右,排升序
然后从右到左,排升序

核心: 用两个数组存储,取其最大值
*/
#include <iostream>
#include <cstring>

using namespace std;
const int N = 10010;
int w[N], milk[N];
int main()
{
    int m;
    cin >> m;
    for(int i = 0; i < m; i++) cin >> w[i];
    milk[0] = 200;
    for(int i = 0; i < m - 1; i ++)
    {
        if (w[i + 1] > w[i]) milk[i + 1] = milk[i] + 100;
        else
        {
            if (w[i + 1] == w[i] || milk[i] == 200) milk[i + 1] = milk[i];
            else milk[i + 1] = milk[i] - 100;
        }
    }
    for(int i = m - 1; i > 0; i--)
    {
        if (w[i] < w[i - 1]) milk[i - 1] = milk[i] + 100;
    }
    int sum = 0;
    for(int i = 0; i < m; i++) sum += milk[i];
    printf("%d", sum);
}