Codefroces432 div2 A,B,C

Posted 2020-10-05 十年换你一句好久不见

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Arpa is researching the Mexican wave.

There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.

At time 1, the first spectator stands.
At time 2, the second spectator stands.
...
At time k, the k-th spectator stands.
At time k?+?1, the (k?+?1)-th spectator stands and the first spectator sits.
At time k?+?2, the (k?+?2)-th spectator stands and the second spectator sits.
...
At time n, the n-th spectator stands and the (n?-?k)-th spectator sits.
At time n?+?1, the (n?+?1?-?k)-th spectator sits.
...
At time n?+?k, the n-th spectator sits.

Arpa wants to know how many spectators are standing at time t.

Input

The first line contains three integers n, k, t (1?≤?n?≤?10⁹, 1?≤?k?≤?n, 1?≤?t?<?n?+?k).

Output

Print single integer: how many spectators are standing at time t.

Examples

Input

10 5 3

Output

Input

10 5 7

Output

Input

10 5 12

Output

Note

In the following a sitting spectator is represented as -, a standing spectator is represented as ^.

At t?=?0? ---------- $技术分享$ number of standing spectators = 0.
At t?=?1? ^--------- $技术分享$ number of standing spectators = 1.
At t?=?2? ^^-------- $技术分享$ number of standing spectators = 2.
At t?=?3? ^^^------- $技术分享$ number of standing spectators = 3.
At t?=?4? ^^^^------ $技术分享$ number of standing spectators = 4.
At t?=?5? ^^^^^----- $技术分享$ number of standing spectators = 5.
At t?=?6? -^^^^^---- $技术分享$ number of standing spectators = 5.
At t?=?7? --^^^^^--- $技术分享$ number of standing spectators = 5.
At t?=?8? ---^^^^^-- $技术分享$ number of standing spectators = 5.
At t?=?9? ----^^^^^- $技术分享$ number of standing spectators = 5.
At t?=?10 -----^^^^^ $技术分享$ number of standing spectators = 5.
At t?=?11 ------^^^^ $技术分享$ number of standing spectators = 4.
At t?=?12 -------^^^ $技术分享$ number of standing spectators = 3.
At t?=?13 --------^^ $技术分享$ number of standing spectators = 2.
At t?=?14 ---------^ $技术分享$ number of standing spectators = 1.
At t?=?15 ---------- $技术分享$ number of standing spectators = 0.

分情况讨论

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int n,m,k;
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    if(k>=0 && k<=m) printf("%d\n",k);
    else if(k>m && k<=n) printf("%d\n",m);
    else if(k>n && k<=n+m) printf("%d\n",m-k+n);
    else printf("0\n");
    return 0;
}

Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a,?b,?c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input

The only line contains six integers a_x,?a_y,?b_x,?b_y,?c_x,?c_y (|a_x|,?|a_y|,?|b_x|,?|b_y|,?|c_x|,?|c_y|?≤?10⁹). It‘s guaranteed that the points are distinct.

Output

Print "Yes" if the problem has a solution, "No" otherwise.

You can print each letter in any case (upper or lower).

　三点共线一定不存在，除此之外，若a到b的距离等于b到c的距离存在。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
struct point
{
    ll x;
    ll y;
}node[4];
ll dist(point a,point b)
{
    return ((a.y-b.y)*(a.y-b.y))+((a.x-b.x)*(a.x-b.x));
}
bool check()
{
    int ans=0;
    if(dist(node[1],node[0])==dist(node[1],node[2])) ans=1;
    if(ans) return true;
    else return false;
}
bool solve()
{
    double a=((node[1].y-node[0].y)*1.0)/((node[1].x-node[0].x)*1.0);
    double b=((node[2].y-node[0].y)*1.0)/((node[2].x-node[0].x)*1.0);
    if(a==b) return true;
    return false;
}
int main()
{
    for(int i=0;i<3;i++)
    {
        scanf("%lld%lld",&node[i].x,&node[i].y);
    }
    if(solve()) puts("No");
    else if(check()) puts("Yes");
    else puts("No");
    return 0;
}

You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two points coincide.

We will call point a bad if there are different points b and c, not equal to a, from the given set such that angle between vectors $技术分享$ and $技术分享$ is acute (i.e. strictly less than $技术分享$ ). Otherwise, the point is called good.

The angle between vectors $技术分享$ and $技术分享$ in 5-dimensional space is defined as $技术分享$ , where $技术分享$ is the scalar product and $技术分享$ is length of $技术分享$ .

Given the list of points, print the indices of the good points in ascending order.

Input

The first line of input contains a single integer n (1?≤?n?≤?10³) — the number of points.

The next n lines of input contain five integers a_i,?b_i,?c_i,?d_i,?e_i (|a_i|,?|b_i|,?|c_i|,?|d_i|,?|e_i|?≤?10³) — the coordinates of the i-th point. All points are distinct.

Output

First, print a single integer k — the number of good points.

Then, print k integers, each on their own line — the indices of the good points in ascending order.

Examples

Input

Output

1
1

Input

Output

Note

In the first sample, the first point forms exactly a $技术分享$ angle with all other pairs of points, so it is good.

In the second sample, along the cd plane, we can see the points look as follows:

$技术分享$

We can see that all angles here are acute, so no points are good.

暴力枚举

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
struct Point{
    ll a, b, c, d, e;
}nod[1006];
bool vis[1006];
ll check(Point aa, Point bb, Point cc) {
    return (bb.a-aa.a)*(cc.a-aa.a)+(bb.b-aa.b)*(cc.b-aa.b)+(bb.c-aa.c)*(cc.c-aa.c)+(bb.d-aa.d)*(cc.d-aa.d)+(bb.e-aa.e)*(cc.e-aa.e);
}
int main(){
    int n;
    memset(vis, true, sizeof(vis));
    cin >> n;
    for(int i = 1; i <= n; i ++) {
        cin >> nod[i].a >> nod[i].b >> nod[i].c >> nod[i].d >> nod[i].e;
    }
    int k = n;
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= n; j ++) {
            for(int l = j+1; l <= n; l ++) {
                if(i != j && i != l) {
                    if(check(nod[i], nod[j], nod[l]) > 0) {
                        vis[i] = false;
                        k--;
                        goto tt;
                    }
                }
            }
        }
        tt:;
    }
    printf("%d\n",k);
    int cnt = 0;
    for(ll i = 1; i <= n; i ++) {
        if(vis[i]) {
            cnt ++;
            printf("%lld ",i);
        }
    }
    if(k!=0)printf("\n");
    return 0;
}

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