笔试强训(二)

一.牛牛的快递

https://www.nowcoder.com/practice/41b42e7b3c3547e3acf8e90c41d98270?tpId=290&tqId=39852&ru=/exam/oj

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    double a;
    char b;
    cin >> a >> b;

    int ret = 0;

    if (a <= 1)
    {
        ret += 20;
    }
    else
    {
        ret += 20;
        a -= 1;
        ret += ceil(a);
    }

    if (b == 'y') ret += 5;

    cout << ret << endl;

    return 0;
}

二.最小花费爬楼梯

https://www.nowcoder.com/practice/9b969a3ec20149e3b870b256ad40844e?tpId=230&tqId=39751&ru=/exam/oj

//楼顶在最后一个数组的后一个位置

//dp[i]表示到达i位置的最小花费(不计算当前位置的cost[i])
#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int cost[N];
int n;
int dp[N];
int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> cost[i];
	}
	for (int i = 2; i <= n; i++)
	{
		dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
	}
	cout << dp[n] << endl;
	return 0;
}

三.数组中两个字符串的最⼩距离

https://www.nowcoder.com/questionTerminal/2c6a0a8e1d20492f92941400036e0890

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main()
{
    int n;
    string s1, s2;
    string s;

    cin >> n;
    cin >> s1 >> s2;

    int prev1 = -1, prev2 = -1, ret = 0x3f3f3f3f;
    for (int i = 0; i < n; i++)
    {
        cin >> s;
        if (s == s1) // 去前面找最近的 s2
        {
            if (prev2 != -1)
            {
                ret = min(ret, i - prev2);
            }
            prev1 = i;
        }
        else if (s == s2) // 去前面找 s1
        {
            if (prev1 != -1)
            {
                ret = min(ret, i - prev1);
            }
            prev2 = i;
        }
    }

    if (ret == 0x3f3f3f3f) cout << -1 << endl;
    else cout << ret << endl;

    return 0;
}