7 28 第二次团队赛 致队友

按照ac题的顺序

H. Making Friends

Ali is trying to make his friends happier by matching them into pairs together. In the beginning, Ali has 2 × n friends standing in a row and numbered from 1 to 2 × n. Each friend i will be matched with the friend numbered (2 × n - i + 1).

Each friend i has a happiness level equal to hi. The happiness level of a matched pair of friends i and (2 × n - i + 1) is equal to hi + h2 × n - i + 1.

Your task is to find the maximum happiness level of a pair among all matched pairs. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 50) specifying the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 1000), giving that Ali has 2 × n friends. Then a line follows containing 2 × n integers h1, ..., h2 × n (1 ≤ hi ≤ 1000), in which hi represents the happiness level of the ith friend.

Output

For each test case, print a single line containing the maximum happiness level of a pair among all matched pairs.

Example

Input

1
3
2 4 5 6 7 8

Output

11

题意   找每次i跟2n-i+1配对的最大值

#include<cstdio>
#include <cstring>
#include <cmath>
#include<algorithm>
#include <stdlib.h>
using namespace std;
int main()
{
    int T,i,n,a[10100];
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=1;i<=2*n;i++)
            scanf("%d",&a[i]);
        int maxx=0;
        for(i=1;i<=n;i++)
        {
           if(a[i]+a[2*n-i+1]>maxx)
                maxx=a[i]+a[2*n-i+1];
        }
        printf("%d\n",maxx);
    }
    return 0;
}

C. Flip the Bits

You are given a positive integer n. Your task is to build a number m by flipping the minimum number of bits in the binary representation of n such that m is less than n (m < n) and it is as maximal as possible. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 105) specifying the number of test cases.

Each test case consists of a single line containing one integer n (1 ≤ n ≤ 109), as described in the statement above.

Output

For each test case, print a single line containing the minimum number of bits you need to flip in the binary representation of n to build the number m.

Example

Input

2
5
10

Output

1
2

题意:有一个n 将2进制中每个位翻转  找到一个比n小且最大的  输出翻转了几位  注意翻转指像硬币一样翻转

思路:找到右边开始第一位是1的位置输出就行  例如二进制的1100  变成1011  3位

AC代码

#include<cstdio>
#include <cstring>
#include <cmath>
#include<algorithm>
#include <stdlib.h>
using namespace std;
int main()
{
    long long int T,i,n,ans=1;
    scanf("%lld",&T);
    while(T--)
    {
        ans=1;
        scanf("%lld",&n);
        while(1)
        {
            if(n%2==1)
                break;
            n/=2;
            ans++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

I. Split the Number

You are given an integer x. Your task is to split the number x into exactly n strictly positive integers such that the difference between the largest and smallest integer among them is as minimal as possible. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.

Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 109, 1 ≤ n ≤ 1000), as described in the statement above.

Output

For each test case, print a single line containing n space-separated integers sorted in a non-decreasing order. If there is no answer, print  - 1.

Example

Input

1
5 3

Output

1 2 2

Note

The strictly positive integers are the set defined as: . That is, all the integers that are strictly greater than zero: .

题意:例如样例 7 4 可以把7分成1 3 3 也可以分成2 2 3 答案是后一种 因为题目要保证最大值与最小值差最小  注意输出-1的情况

#include<cstdio>
#include <cstring>
#include <cmath>
#include<algorithm>
#include <stdlib.h>
using namespace std;
int main()
{
    long long int T,x,n,a[10100],i;
    scanf("%lld",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%lld%lld",&x,&n);
        if(x<n)
            printf("-1\n");
        else
        {
            long long int p,q;
            p=x/n;
            q=x%n;
            for(i=0; i<n; i++)
                a[i]+=p;
            long long  int d=0;
            for(i=n-1; i>=0; i--)
            {
                if(d==q)
                    break;
                d++;
                a[i]+=1;
            }
            for(i=0; i<n; i++)
            {
                if(i==n-1)
                    printf("%lld\n",a[i]);
                else
                    printf("%lld ",a[i]);
            }
        }
    }
    return 0;
}

B. Friends and Cookies

Abood's birthday has come, and his n friends are aligned in a single line from 1 to n, waiting for their cookies, Abood has x cookies to give to his friends.

Here is an example to understand how Abood gives away the cookies. Suppose Abood has 4 friends and x cookies, then Abood will do the following:

  1. Give a cookie to the 1st friend.
  2. Give a cookie to the 2nd friend.
  3. Give a cookie to the 3rd friend.
  4. Give a cookie to the 4th friend.
  5. Give a cookie to the 3rd friend.
  6. Give a cookie to the 2nd friend.
  7. Give a cookie to the 1st friend.
  8. Give a cookie to the 2nd friend.
  9. And so on until all the x cookies are given away.

Your task is to find how many cookies each friend will get. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.

Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 1018, 1 ≤ n ≤ 1000), in which x is the number of cookies Abood has, and n is the number of his friends.

Output

For each test case, print a single line containing n space-separated integers a1, ..., an, in which ai represents how many cookies the ith friend got.

Example

Input

1
5 3

Output

2 2 1

题意:分蛋糕  就像下表一样  输出没人分得的蛋糕数

假设有5个人  10块蛋糕  看下表

A B C D E
1 2 3 4 5
9 8 7 6  
  10      

 

 

 

 

 

所以最后输出 2 3 2 2 1

。。。。这个题交了7遍  一直都是第三组运行错误  就在最后终于找到了错误  要特判一下n=1的情况

下面是ac代码

#include<cstdio>
#include <cstring>
#include <cmath>
#include<algorithm>
#include <stdlib.h>
using namespace std;
int main()
{
    long long int T,m,n,a[101000],i;
    scanf("%lld",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%lld%lld",&m,&n);
        if(n==1)
        {
            printf("%lld\n",m);
        }
        else
        if(m<n)
        {
            for(i=0;i<m;i++)
               a[i]=1;
            for(i=0;i<n;i++)
            {
                if(i==n-1)
                    printf("%lld\n",a[i]);
                else
                    printf("%lld ",a[i]);
            }
        }
        else
        {
            long long x,y;
            for(i=0;i<n;i++)
                a[i]=1;
            m=m-n;
            x=m/(n-1);
            y=m%(n-1);
           for(i=1;i<n-1;i++)
              a[i]+=x;
            if(x%2!=0)
            {
                a[0]+=x/2+1;
                a[n-1]+=x/2;
                for(i=1;i<=y;i++)
                    a[i]+=1;
            }
            else
            {
                a[0]+=x/2;
                a[n-1]+=x/2;
                long long int f=0;
                for(i=n-2;i>=1;i--)
                {
                    if(f>=y)
                        break;
                    a[i]++;
                    f++;
                }
            }
            for(i=0;i<n;i++)
            {
                if(i==n-1)
                    printf("%lld\n",a[i]);
                else
                    printf("%lld ",a[i]);
            }
        }
    }
    return 0;
}

这次较上次有进步了  继续努力~~~~

posted @ 2018-07-28 19:48  ~~zcy  阅读(101)  评论(0编辑  收藏  举报