I题

Sample Input
4 2 6 8 
5
1
结尾无空行
Sample Output
6 4 
结尾无空行
Acknowledgment
2012 Laboratories of HUAWEI.

题目大意：

在一行中给出一个集合中的元素，输入一个目标值A和半径r，输出和A的距离小于等于r的元素（降序输出），如果没有符合条件的点输出空行。

坑点：

输入的第一行每个元素后面都有空格，如果用getchar判断会WA

用getline输入可以避免上述问题

#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
#include<map>
#include<stack>
#include<cstdio>
#include<cstring>
#define ll long long
#define ull unsigned long long
using namespace std;
const int N = 1e8 + 10;
bool cmp(int a, int b)
{
    return a > b;
}
int a[N];
string s;
int main()
{
    int x;
    getline(cin, s);
    int n = 0;
    int num = 0;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] == ' ')
        {
            a[n++] = num;
            num = 0;
        }
        else
        {
            num = num * 10 + s[i] - '0';
        }
    }
    int o, r;
    cin >> o >> r;
    bool f = false;
    sort(a, a + n, cmp);
    for (int i = 0; i < n; i++)
    {
        if (abs(a[i] - o) <= r)
        {
            f = true;
            cout << a[i] << ' ';
        }
    }
     if (!f)
         cout << endl;
    return 0;
}

F题

In the sample input, we have Chasm at (3,5), Jueyun Karst at (6,0), and Keqing can finish the visiting within radius 1.

So Keqing could go directly to point (3,4) first to visit Chasm, then reach (5.4,0.8) to visit Jueyun Chasm. The total length is 5+4=9. We can prove that there is no path with a smaller distance.

Sample Input
1
3 5 1
结尾无空行
Sample Output
Case #1: 9.00
结尾无空行
Acknowledgment
ICT Products & Solutions of HUAWEI.

题目大意：

从原点出发先到A(a,b)再去B(2a,0)。求最短总路程。(距离某点小于R时，即视为到达该点)。

分析：

如果r<b，先经过以A为圆心，r为半径的圆上距离x轴最近的点（记为P），再去PB与以B为圆心，r为半径的圆的交点。

否则直接沿x轴通过，总路程最短。

坑点：

a, b, r如果定义为int会因为溢出而WA，改为long long即可。

#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
#include<map>
#include<stack>
#include<cstdio>
#include<cstring>
#define ll long long
#define ull unsigned long long
using namespace std;
int main()
{
    int t;
    int flag = 1;

    scanf("%d", &t);

    while (t--)
    {
        ll a, b, r;

        cin >> a >> b >> r;

        if (r > b)
        {
            printf("Case #%d: %.2f", flag++, 2.0 * a - r);
        }
        else
        {
            printf("Case #%d: %.2f", flag++, 2.0 * sqrt(a * a + (b - r) * (b - r)) - r);
        }
        if (t) puts("");
    }


    return 0;
}