9.6训练日志

A - Arpa and a research in Mexican wave

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.
代码分三类情况讨论一下即可`

#include<bits/stdc++.h>using namespace std;int main(){int n,k,t,i;cin>>n>>k>>t;if(t<=k) {cout<<t; }else if(t>k&&t<=n) {cout<<k;  } else if(t>n){cout<<k+n-t;  }  }

B - Arpa and an exam about geometry

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.
判断是否共线以及ab，bc距离是否相等即可

#include<bits/stdc++.h>using namespace std;long long int f(long long int a,long long int b,long long int c,long long int d) {    return (c-a)*(c-a)+(d-b)*(d-b);}int main() {    long long int ax,ay,bx,by,cx,cy;    cin>>ax>>ay>>bx>>by>>cx>>cy;    if(f(ax,ay,bx,by)==f(bx,by,cx,cy)&&(bx-ax)*(by-cy)!=(bx-cx)*(by-ay)) {        cout<<"YES";        return 0;    }    cout<<"NO";    return 0;}

C - Five Dimensional Points

五维空间算角度，最多11个相邻，之后用公式套即可

#include<bits/stdc++.h>using namespace std;long long int m,a[120][6];int f(int x){long long int i,j,k,n,p;for(i=1;i<=m; i++)  for(j=i+1;j<=m;j++)     if(x!=i&&x!=j)      {n=0;p=0;       for(k=1;k<=5;k++)        n+=(a[j][k]-a[x][k])*(a[i][k]-a[x][k]);       if(n>0)return 0;      }      return 1;      }int main() {    long long int i,j,x,n,k,ans=0,b[120]={0};bool p=0;    cin>>m;    if(m>11) {        cout<<0;        return 0;    }    for(i=1;i<=m;i++)        for(j=1;j<=5;j++)            cin>>a[i][j];   for(x=1;x<=m;x++)        {      if(f(x)==1)      {b[x]=1;ans++;      }    }     cout<<ans<<endl;     for(i=1;i<=m;i++)     if(b[i]==1)cout<<i<<endl;    }

D - Polycarpus’ Dice

Polycarp has n dice d1, d2, …, dn. The i-th dice shows numbers from 1 to di. Polycarp rolled all the dice and the sum of numbers they showed is A. Agrippina didn’t see which dice showed what number, she knows only the sum A and the values d1, d2, …, dn. However, she finds it enough to make a series of statements of the following type: dice i couldn’t show number r. For example, if Polycarp had two six-faced dice and the total sum is A = 11, then Agrippina can state that each of the two dice couldn’t show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible).

For each dice find the number of values for which it can be guaranteed that the dice couldn’t show these values if the sum of the shown values is A.
每一次判断可取种数，是否加上剩余数的和大于所求数，是否小于所求数减去剩余数数量

#include<bits/stdc++.h>using namespace std;const int N=200009;int main() {    long long int A=0,p,x,n,i,a[N],ans=0;    cin>>n>>A;    for(i=1; i<=n; i++) {        cin>>a[i];        ans+=a[i];    }    if(n==1)        cout<<a[1]-1;    else        for(i=1; i<=n; i++) {            x=0;            p=ans-a[i];            p=A-p;            if(p>1)x=p-1;            if(a[i]+n-1>A)                x=x-A+a[i]+n-1;            if(a[i]-1<=x)             cout<<a[i]-1<<" ";            else cout<<x<<" ";        }}