Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 730 Accepted Submission(s): 341

开根向下取整；
Problem Description

Let's define the function

f(n)=⌊n−−√⌋.

Bo wanted to know the minimum number

y which satisfies

fy(n)=1.

note:

f1(n)=f(n),fy(n)=f(fy−1(n))

It is a pity that Bo can only use 1 unit of time to calculate this function each time.

And Bo is impatient, he cannot stand waiting for longer than 5 units of time.

So Bo wants to know if he can solve this problem in 5 units of time.

Input

This problem has multi test cases(no more than

120).

Each test case contains a non-negative integer

n(n<10100).

Output

For each test case print a integer - the answer

y or a string "TAT" - Bo can't solve this problem.

Sample Input

    233 233333333333333333333333333333333333333333333333333333333 
  

Sample Output

    3 TAT 
  

Source

2016 Multi-University Training Contest 3

Recommend

wange2014 | We have carefully selected several similar problems for you: 5761 5760 5759 5758 5757

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define LL long long
using namespace std;
char s[210];
LL ans;
LL A[]={1,(1<<2)-1,(1<<4)-1,(1<<8)-1,(1<<16)-1,(1LL<<32)-1};
int main()
{
    //cout<<A[5]<<endl;
    while(~scanf("%s",s))
    {
        if(strlen(s)>10||strlen(s)==1&&s[0]=='0')
            printf("TAT\n");
        else
        {
            ans=0;

            for(int i=0;s[i];i++)
                ans=ans*10+s[i]-'0';
                if(ans>A[5])
                    printf("TAT\n");
                else
            printf("%d\n",lower_bound(A,A+6,ans)-A);
        }
    }
    return 0;
}

Permutation Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 508 Accepted Submission(s): 301
Special Judge

求系数；
Problem Description

There are two sequences

h1∼hn and

c1∼cn.

h1∼hn is a permutation of

1∼n. particularly,

h0=hn+1=0.

We define the expression

[condition] is 1 when

condition is True,is 0 when

condition is False.

Define the function

f(h)=∑ni=1ci[hi>hi−1 and hi>hi+1]

Bo have gotten the value of

c1∼cn, and he wants to know the expected value of

f(h).

Input

This problem has multi test cases(no more than

12).

For each test case, the first line contains a non-negative integer

n(1≤n≤1000), second line contains

n non-negative integer

ci(0≤ci≤1000).

Output

For each test cases print a decimal - the expectation of

f(h).

If the absolute error between your answer and the standard answer is no more than

10−4, your solution will be accepted.

Sample Input

     4 3 2 4 5 5 3 5 99 32 12 
   

Sample Output

     6.000000 52.833333 
   

Source

2016 Multi-University Training Contest 3

Recommend

wange2014 | We have carefully selected several similar problems for you: 5761 5760 5759 5758 5757

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
LL A[1010]={0};
LL sum[1010]={0};
int main()
{
    int n;
     for(int i=2;i<1010;i++)
        A[i]=(LL)(i-1)*(i-2)*(i);
    LL ans1,ans2;
    int x;
    LL ans=0;
        for(int i=3;i<1010;i++)
        {
            ans+=(LL)(i-2);
            sum[i]=sum[i-1]+ans*2;
        }
    while(~scanf("%d",&n))
    {
        ans1=ans2=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(i!=n&&i!=1)
                ans2+=x;
            else
                ans1+=x;
        }
        if(n==1)
        {
            printf("%.10f\n",(double)ans1);
            continue;
        }
        else
        {
            double tmp=(double)ans1/2.0;
            if(n>2)
                tmp+=+ans2*(sum[n])/(double)(A[n]);
            printf("%.10f\n",tmp);
        }
    }
    return 0;
}

Teacher Bo

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 691 Accepted Submission(s): 381

循环不会超过2*M；
Problem Description

Teacher BoBo is a geography teacher in the school.One day in his class,he marked

N points in the map,the

i-th point is at

(Xi,Yi).He wonders,whether there is a tetrad

(A,B,C,D)(A<B,C<D,A≠CorB≠D) such that the manhattan distance between A and B is equal to the manhattan distance between C and D.

If there exists such tetrad,print "YES",else print "NO".

Input

First line, an integer

T. There are

T test cases.

(T≤50)

In each test case,the first line contains two intergers, N, M, means the number of points and the range of the coordinates.

(N,M≤105).

Next N lines, the

i-th line shows the coordinate of the

i-th point.

(Xi,Yi)(0≤Xi,Yi≤M).

Output

T lines, each line is "YES" or "NO".

Sample Input

     2 3 10 1 1 2 2 3 3 4 10 8 8 2 3 3 3 4 4 
   

Sample Output

     YES NO 
   

Source

2016 Multi-University Training Contest 3

Recommend

wange2014 | We have carefully selected several similar problems for you: 5761 5760 5759 5758 5757

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
#define LL long long
using namespace std;
const int N=1e5+10;
struct node
{
    int x,y;
    bool operator <(const node &A)const
    {
        return (x==A.x&&y<A.y||x<A.x)?1:0;
    }
    bool operator ==(const node &A)const
    {
        return (x==A.x&&y==A.y)?1:0;
    }
}que[N];
int dis[N<<2];
 int n,m;
 bool slove()
 {
     memset(dis,0,sizeof(dis));
     for(int i=0;i<n;i++)
     {
         for(int j=i+1;j<n;j++)
         {
             int d=abs(que[i].x-que[j].x)+abs(que[i].y-que[j].y);
             dis[d]++;
             if(dis[d]>1)
                return 1;
         }
     }
     return 0;
 }
int main()
{

    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&que[i].x,&que[i].y);
        }
//        if(sqrt(m)<=n)
//        {
//            printf("YES\n");
//            continue;
//        }
        sort(que,que+n);
        n=unique(que,que+n)-que;
        if(slove())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

2016 Multi-University Training Contest 3

Sqrt Bo

Permutation Bo

Teacher Bo

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