Codeforces Round #196 (Div. 2)

The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her n students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).

The shop assistant told the teacher that there are m puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of f₁ pieces, the second one consists of f₂ pieces and so on.

Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let A be the number of pieces in the largest puzzle that the teacher buys and B be the number of pieces in the smallest such puzzle. She wants to choose such n puzzles that A - B is minimum possible. Help the teacher and find the least possible value of A - B.

Input

The first line contains space-separated integers n and m (2 ≤ n ≤ m ≤ 50). The second line contains m space-separated integersf₁, f₂, ..., f_m (4 ≤ f_i ≤ 1000) — the quantities of pieces in the puzzles sold in the shop.

Output

Print a single integer — the least possible difference the teacher can obtain.

        Sample test(s) 
      
          input 
        
4 6
10 12 10 7 5 22

          output 
        
5

Note

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

水题。排序枚举。

#include <iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[60],n,m,ans;
int main()
{
    int i,j;

    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<m;i++)
            scanf("%d",&a[i]);
        sort(a,a+m);
        ans=0x3f3f3f3f;
        for(i=0;i+n-1<m;i++)
        {
            if(a[i+n-1]-a[i]<ans)
                ans=a[i+n-1]-a[i];
        }
        printf("%d\n",ans);
    }
    return 0;
}

Manao has a monitor. The screen of the monitor has horizontal to vertical length ratio a:b. Now he is going to watch a movie. The movie's frame has horizontal to vertical length ratio c:d. Manao adjusts the view in such a way that the movie preserves the original frame ratio, but also occupies as much space on the screen as possible and fits within it completely. Thus, he may have to zoom the movie in or out, but Manao will always change the frame proportionally in both dimensions.

Calculate the ratio of empty screen (the part of the screen not occupied by the movie) to the total screen size. Print the answer as an irreducible fraction p / q.

Input

A single line contains four space-separated integers a, b, c, d (1 ≤ a, b, c, d ≤ 1000).

Output

Print the answer to the problem as "p/q", where p is a non-negative integer, q is a positive integer and numbers p and q don't have a common divisor larger than 1.

        Sample test(s) 
      
          input 
        
1 1 3 2

          output 
        
1/3

          input 
        
4 3 2 2

          output 
        
1/4

Note

Sample 1. Manao's monitor has a square screen. The movie has 3:2 horizontal to vertical length ratio. Obviously, the movie occupies most of the screen if the width of the picture coincides with the width of the screen. In this case, only 2/3 of the monitor will project the movie in the horizontal dimension:

Sample 2. This time the monitor's width is 4/3 times larger than its height and the movie's frame is square. In this case, the picture must take up the whole monitor in the vertical dimension and only 3/4 in the horizontal dimension:

只可能出现两种情况。即先把图像的长变为和框架的长一样或把图像的高变为和框架的高一样。两种情况比例分别为

(b*c-a*d)/(b*c)和(a*d-b*c)/(a*d)。因为b*c-a*d和a*d-b*c互为相反数所以取正的那个就行了。只是要注意答案要约分。还有分子为0的情况。分母必须为1。所以要特殊处理。

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

int a,b,c,d;
int gcd(int x,int y)//最大公约数
{
    int t;
    while(y)
    {
        t=x%y;
        x=y;
        y=t;
    }
    return x<0?-x:x;//注意正负
}
int main()
{
    int p,q,e,f,t,ans,mo;

    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        p=a*d-b*c;
        q=a*d;
        e=b*c-a*d;
        f=b*c;
        if(p==0||e==0)
        {
            printf("0/1\n");
            continue;
        }
        t=gcd(p,q);
        p/=t;
        q/=t;
        t=gcd(e,f);
        e/=t;
        f/=t;
        if(p>0)
            ans=p,mo=q;
        else
            ans=e,mo=f;
        printf("%d/%d\n",ans,mo);
    }
    return 0;
}

Manao is taking part in a quiz. The quiz consists of n consecutive questions. A correct answer gives one point to the player. The game also has a counter of consecutive correct answers. When the player answers a question correctly, the number on this counter increases by 1. If the player answers a question incorrectly, the counter is reset, that is, the number on it reduces to 0. If after an answer the counter reaches the number k, then it is reset, and the player's score is doubled. Note that in this case, first 1 point is added to the player's score, and then the total score is doubled. At the beginning of the game, both the player's score and the counter of consecutive correct answers are set to zero.

Manao remembers that he has answered exactly m questions correctly. But he does not remember the order in which the questions came. He's trying to figure out what his minimum score may be. Help him and compute the remainder of the corresponding number after division by 1000000009 (10⁹ + 9).

Input

The single line contains three space-separated integers n, m and k (2 ≤ k ≤ n ≤ 10⁹; 0 ≤ m ≤ n).

Output

Print a single integer — the remainder from division of Manao's minimum possible score in the quiz by 1000000009 (10⁹ + 9).

        Sample test(s) 
      
          input 
        
5 3 2

          output 
        
3

          input 
        
5 4 2

          output 
        
6

Note

Sample 1. Manao answered 3 questions out of 5, and his score would double for each two consecutive correct answers. If Manao had answered the first, third and fifth questions, he would have scored as much as 3 points.

Sample 2. Now Manao answered 4 questions. The minimum possible score is obtained when the only wrong answer is to the question 4.

Also note that you are asked to minimize the score and not the remainder of the score modulo 1000000009. For example, if Manao could obtain either 2000000000 or 2000000020 points, the answer is 2000000000 mod 1000000009, even though2000000020 mod 1000000009 is a smaller number.

题目意思很简单。就说一人去参加比赛。每答对一题得一分。连续答对k题时先加一分然后总分数乘2然后连续答对题数归0。然后给你题目总数n和他答对题目的个数m问你他能得的最低分是多少。

这题关键是找出最优方案(使得分最低)。

我们设他答错w题。如果w*k>=n。那么他的得分就为n-w。因为这w次错的可以使他答题连续对的小于k个。应该很好理解。

如果w*k<n说明会出现连续答对超过n个的情况。既然这种情况不能避免那么尽量让它出现在前面。因为前面分数少乘以2对总分的影响比较小。所以分数可以分为两部分。

1，前面连续的部分。2，后面被分隔的部分。

后面部分的分数很好求为：(k-1)*w。

而前面的分数也是有规律的。我们假设前面有n*k次答对。

那么分数变化为：2*k,(2*k+k)*2,((2*k+k)*2+k)*2......。抽象出来这就是一个数列递推公式为：

an=2*(an-1+k)。

展开移项后得：an+2*k=2*(an-1+2*k)。设cn=an+2*k。那么cn/cn-1=2。等比数列。

而c0=a0+2*k=2k。那么cn=2*k*2^n。所以an=k*2^(n+1)-2*k。

所以前面的得分为：

k*2^((n-w*k)/k+1)-2*k+(n-w*k)%k。

所以总分最小值就出来了。

ans=k*2^((n-w*k)/k+1)-2*k+(n-w*k)%k+(k-1)*w。

详细见代码：

#include <iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod=1000000009;
ll pow_mod(ll a,ll i)//快速幂
{
    if(i==0)
        return 1%mod;
    ll temp=pow_mod(a,i>>1);//这里也必须是ll
    temp=temp*temp%mod;
    if(i&1)
        temp=(ll)temp*a%mod;
    return temp;
}
ll n,m,w,ans,t,k;
int main()
{
    while(~scanf("%I64d%I64d%I64d",&n,&m,&k))
    {
        w=n-m;
        t=n-k*w;
        if(t<=0)
        {
            printf("%I64d\n",(n-w)%mod);
            continue;
        }
        ans=(k-1)*w%mod;
        t=(pow_mod(2,t/k+1)*k%mod+mod-2*k%mod+t%k%mod)%mod;//注意减时可能减为负数所以要加上mod因此听取蛙声一片
        ans=(t+ans)%mod;
        printf("%I64d\n",ans);
    }
    return 0;
}

决定以后好好打cf了。未完待续。。。。

Codeforces Round #196 (Div. 2)

你可能感兴趣的:(c,算法,ACM)