CodeForces Round #185 (Div. 2)A,B,C

水平稀，只能做A，B，C

A. Whose sentence is it?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.

Input

The first line of the input contains an integer n (1≤n≤10), number of sentences in the chat record. Each of the next n lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.

Output

For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.

Sample test(s)

input

5
I will go to play with you lala.
wow, welcome.
miao.lala.
miao.
miao .

output

Freda's
OMG>.< I don't know!
OMG>.< I don't know!
Rainbow's
OMG>.< I don't know!

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

int main()

{

    int T;

    cin>>T;

    getchar();

while(T--)

{

    char str[111];

    gets(str);

    int F=0,R=0;

    int len=strlen(str);

 //   for(int i=0;i<len;i++)

 //       cout<<i<<":"<<str<<" ";

 //   cout<<endl;

    if(str[0]=='m'&&str[1]=='i'&&str[2]=='a'&&str[3]=='o'&&str[4]=='.')

        R=1;

    if(str[len-1]=='.'&&str[len-2]=='a'&&str[len-3]=='l'&&str[len-4]=='a'&&str[len-5]=='l')

        F=1;

 //   cout<<R<<",,,,"<<F<<endl;

    if(F==1&&R==0)

        cout<<"Freda's"<<endl;

    else if(R==1&&F==0)

        cout<<"Rainbow's"<<endl;

    else

        cout<<"OMG>.< I don't know!"<<endl;

}

    return 0;

}

B. Archer

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is  for SmallR while  for Zanoes. The one who shoots in the target first should be the winner.

Output the probability that SmallR will win the match.

Input

A single line contains four integers .

Output

Print a single real number, the probability that SmallR will win the match.

The answer will be considered correct if the absolute or relative error doesn't exceed 10-6.

Sample test(s)

input

1 2 1 2

output

0.666666666667

#include <iostream>

#include <cstdio>

using namespace std;

int main()

{

    double a,b,c,d;

    cin>>a>>b>>c>>d;

    double p1=a/b;

    double p2=c/d;

    double Q=(1-p1)*(1-p2);

    int T=30;

    double sum=p1*(1/(1-Q));

    printf("%.10lf\n",sum);

    return 0;

}

 

C. The Closest Pair

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.

The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1,y1) and (x2,y2) is .

The pseudo code of the unexpected code is as follows:

input n
for i from 1 to n
input the i-th point's coordinates into p
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p.x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p,p[j]))
output d

Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.

You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?

Input

A single line which contains two space-separated integers n and k (2≤n≤2000, 1≤k≤109).

Output

If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi,yi (|xi|,|yi|≤109)representing the coordinates of the i-th point.

The conditions below must be held:

• All the points must be distinct.
• |xi|,|yi|≤109.
• After running the given code, the value of tot should be larger than k.

Sample test(s)

input

4 3

output

0 0
0 1
1 0
1 1

input

2 100

output

no solution

 
 #include <iostream>

#include <cmath>

#include <cstdio>

using namespace std;

int main()

{

    int n;unsigned long long int k;

    cin>>n>>k;

    unsigned long long int tot=0;

    for(int i=1;i<=n;i++)

        for(int j=i+1;j<=n;j++)

          tot++;

    if(tot<=k)  cout<<"no solution"<<endl;

    else

    {

        for(int i=1;i<=n;i++)

            printf("0 %d\n",i);

    }

    return 0;

}