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====== AtCoder Beginner Contest 174 ====== ===== A-Air Conditioner ===== 水题，水到直接在网页的提交代码框里写了。略 ===== B Distance ===== **题意**：有多少个点距离原点小于d。


#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long  ll;
int main()
{
    ll n,d;
    cin >> n >> d;
    d *= d;
    int ans = 0;
    for (int i = 1; i <= n;i++)
    {
        ll x, y;
        cin >> x >> y;
        if (x*x + y*y <= d)
            ans++;
    }
    cout << ans << endl;
    return 0;
}

===== C Repsept ===== **题意**：给出一个正整数 K，问在全部由7组成的数字中，求出一个最短的数字可以被K整除，如果没有就输出-1。 **思路**：K <= 1e6,所以直接暴力计算模数即可，同时记录模数，出现循环，就不存在，输出-1.


#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long  ll;
int vztd[5000000];
int main()
{
    ll k;
    cin >> k;
    ll tem = 7;
    ll ans = 1;
    while (tem < k)
        tem = tem * 10 + 7,ans++;
    for (;;)
    {
        if (tem % k == 0)
        {
            cout << ans << endl;
            return 0;
        }
        ans++;
        tem = tem * 10 + 7;
        tem %= k;
        if (vztd[tem] == 1)
            break;
        else
            vztd[tem] = 1;
    }
    cout << -1 << endl;
    return 0;
}

##　D Alter Altar 水题暴力前后枚举，遇到相反就交换。


#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long  ll;
char s[200009];
int main()
{
    int n;
    cin >> n;
    scanf("%s", s+1);
    int ans = 0, l = 1, r = n;
    while (l= r)
            break;
        ans++;
        swap(s[l], s[r]);
    }
    cout << ans << endl;
    return 0;
}

===== E Logs ===== **题意：**给出一堆原木，每次可以挑一根砍成两段，问砍k次，之后要求最长的原木的最小值。 **题解：**看到这个最大值最小就应该想到二分答案，直接二分即可，注意round up 是向上取整，，多个翻译软件都会翻译成四舍五入。


#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAX = 2e5 + 20;
typedef long long  ll;
int pic[MAX],n,k;
bool check(int goal)
{
    int tem = k,i;
    for (i  = 1; i <= n;i++)
    {   
        if (pic[i] <= goal)
            continue;
        else
        {
            int ans = 0,logg = pic[i];
            ans = logg / goal;
            if (logg % goal == 0)
                ans--;
            tem -= ans;
            if (tem < 0)
                return false;
        }
    }
    return true;
}
int main()
{
    cin >> n >> k;
    int l = 1, r = 0;
    for (int i = 1; i <= n;i++)
    {
        scanf("%d", &pic[i]);
        r = max(pic[i], r);
    }
    while (l + 2 < r)
    {
        int mid = (l + r) >> 1;
        if (check(mid))
            r = mid;
        else
            l = mid + 1;
    }
    for (int i = l; i <= r;i++)
        if (check(i))
        {
            cout << i << endl;
            break;
        }
    return 0;
}

===== F Range Set Query ===== **题意**：静态区间求种类数。经典题了 **题解**：树状数组离线，主席树在线都可以水过。


#include 
#include 
#include 
using namespace std;
struct data
{
    int z, y, xh;
    bool operator<(const data &x) const
    {
        if (y < x.y)
            return 1;
        else
            return 0;
    }
} s[501001];
int c[500011];
int a[500101];
int h[500101];

map Map;
int n;
int lowbit(int i)
{
    return i & (-i);
}
void add(int i, int x)
{
    while (i <= n)
    {
        c[i] += x;
        i += lowbit(i);
    }
}
int sum(int x)
{
    int ans = 0;
    while (x > 0)
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(0);
    int m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];

    for (int i = 1; i <= m; ++i)
    {
        cin >> s[i].z >> s[i].y;
        s[i].xh = i;
    }

    sort(s + 1, s + 1 + m);
    int last = 0;
    for (int i = 1; i <= m; ++i)
    {
        ++last;
        for (int j = last; j <= s[i].y; ++j)
        {
            if (Map[a[j]])
            {
                add(Map[a[j]], -1);
                Map[a[j]] = j;
                add(Map[a[j]], 1);
            }
            else
            {
                Map[a[j]] = j;
                add(j, 1);
            }
        }
        h[s[i].xh] = sum(s[i].y) - sum(s[i].z - 1);
        last = s[i].y;
    }
    for (int i = 1; i <= m; ++i)
        cout << h[i] << endl;
}