cf424（A，B，C）

A. Squats

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.

For another exercise, Pasha needs exactly  hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?

Input

The first line contains integer n (2 ≤ n ≤ 200; n is even). The next line contains n characters without spaces. These characters describe the hamsters' position: the i-th character equals 'X', if the i-th hamster in the row is standing, and 'x', if he is sitting.

Output

In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.

Sample test(s)

input

4
xxXx

output

1
XxXx

input

2
XX

output

1
xX

input

6
xXXxXx

output

0
xXXxXx

简单模拟

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
char a[205];

int main()
{
    int n,s=0;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        if(a[i]=='X')
            s++;
    }
    if(s==n/2)
        cout<<0<<endl<<a<<endl;
    else
    {
        cout<<abs(s-n/2)<<endl;
        if(s<n/2)
        {
            s=n/2-s;
            for(int i=0; i<n; i++)
                if(s&&a[i]=='x')
                    cout<<'X',s--;
               else cout<<a[i];
        }
        else
        {
            s=s-n/2;
            for(int i=0; i<n; i++)
                if(s&&a[i]=='X')
                    cout<<'x',s--;
                else cout<<a[i];
        }
    }
    return 0;
}

B. Megacity

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.

The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with n other locations, the i-th one has coordinates (xi, yi) with the population of ki people. You can widen the city boundaries to a circle of radius r. In such case all locations inside the circle and on its border are included into the city.

Your goal is to write a program that will determine the minimum radius r, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.

Input

The first line of the input contains two integers n and s (1 ≤ n ≤ 103; 1 ≤ s < 106) — the number of locatons around Tomsk city and the population of the city. Then n lines follow. The i-th line contains three integers — the xi and yi coordinate values of the i-th location and the number ki of people in it (1 ≤ ki < 106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.

It is guaranteed that no two locations are at the same point and no location is at point (0; 0).

Output

In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.

The answer is considered correct if the absolute or relative error don't exceed 10 - 6.

Sample test(s)

input

4 999998
1 1 1
2 2 1
3 3 1
2 -2 1

output

2.8284271

input

4 999998
1 1 2
2 2 1
3 3 1
2 -2 1

output

1.4142136

input

2 1
1 1 999997
2 2 1

output

-1

读懂题意挺坑的。。。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

struct data
{
    int x,y,k;
    double r;
}a[10010];

bool cmp(data a,data b)
{
    return a.r<b.r;
}

int main()
{
    int n,s;
    cin>>n>>s;
    for(int i=0;i<n;i++)
        cin>>a[i].x>>a[i].y>>a[i].k,a[i].r=sqrt((double)a[i].x*(double)a[i].x+(double)a[i].y*(double)a[i].y);
    sort(a,a+n,cmp);
    int flag=0;double rr;
    for(int i=0;i<n;i++)
    {
        //cout<<a[i].r<<endl;
        s+=a[i].k;
        if(s>=1000000){flag=1;rr=a[i].r;break;}
    }
    if(flag)printf("%.7f\n",rr);
    else cout<<-1<<endl;
    return 0;
}

C. Magic Formulas

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

People in the Tomskaya region like magic formulas very much. You can see some of them below.

Imagine you are given a sequence of positive integer numbers p1, p2, ..., pn. Lets write down some magic formulas:

Here, "mod" means the operation of taking the residue after dividing.

The expression  means applying the bitwise xor (excluding "OR") operation to integers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by "^", in Pascal — by "xor".

People in the Tomskaya region like magic formulas very much, but they don't like to calculate them! Therefore you are given the sequence p, calculate the value of Q.

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 106). The next line contains n integers: p1, p2, ..., pn (0 ≤ pi ≤ 2·109).

Output

The only line of output should contain a single integer — the value of Q.

Sample test(s)

input

3
1 2 3

output

3

找规律题，有点儿递推的意思~不过，话说scanf真的很快，才跑了249ms，不知为何用cin却2000ms都跑不完。。。。。。不是说cin只是scanf的3倍吗。。。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;

ll yh[1000010]={0};

int main()
{
    ll n;
    cin>>n;
    ll ans=0;
    for(ll i=0;i<n;i++)
    {
        yh[i+1]=(i+1)^yh[i];
        ll tem;
        scanf("%lld",&tem);
        ans^=tem;
        ans^=yh[i];
        if(i>1)
        {
            ll x=(n-i)/i;
            ll sh=n-i-x*i;
            ans^=yh[sh];
            if(x%2==1)
                ans^=yh[i-1];
        }
    }
    cout<<ans<<endl;
    return 0;
}