BestCoder Round #11 (Div. 2) Solutions to the first three questions_html/css_WEB-ITnose

                  <p class="sycode">                      10 10 5 510 10 6 6                  </p>

                  <p class="sycode">                      YESNO                  </p>

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<vector>#include<cmath>#include<string>#include<queue>#define eps 1e-9#define ll long long#define INF 0x3f3f3f3fusing namespace std;int  main(){    int x,y,n,m;    while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF)    {        if(2*x==n&&2*y==m)            printf("YES\n");        else            printf("NO\n");    }  } <br> hdu 5055 Bob and math problem </p> <p class="sycode"> Question meaning : </p> <p class="sycode"> is to give n numbers, and then ask you to find a number that meets the following conditions.  </p> <p class="sycode"> (1) This number is an odd number.  </p> <p class="sycode"> (2) This number is the largest number.  </p> <p class="sycode"> (3) Another point that has been challenged is that all numbers must be used. I was just 0 0 1 beaten. . I also deliberately judged this situation because I didn't understand the question. .  </p> <p class="sycode"> Idea: </p> <p class="sycode"> The greedy approach is to first check whether there is a base in all the bits. If there is no base, then there must be no such number. Secondly, if there is, then the minimum Find out the base of each bit, then sort all the bits, and then assign values ​​from low bits to high bits, then you will get this tree. Finally, judge, if the first bit is 0, then this number does not exist, because it is required to output all bit. .  </p> <p class="sycode"> Question: </p> <p class="sycode"> Bob and math problem <strong>Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)<br> Total Submission (s): 643 Accepted Submission(s): 245<br> </strong> <br> <br> </p>
<p class="sycode"> Problem Description </p>   <p class="sycode">    Recently, Bob has been thinking about a math problem.    <br> There are N Digits, each digit is between 0 and 9. You need to use this N Digits to constitute an Integer.    <br> This Integer needs to satisfy the following conditions:    <br>    </p>
<li>1. must be an odd Integer.<br> </li>    <li>2. there is no leading zero.<br> </li>    <li>3. find the biggest one which is satisfied 1, 2.</li>    <br> Example:     <br> There are three Digits: 0, 1, 3. It can constitute six number of Integers. Only "301", "103" is legal, while "130", "310", "013", "031" is illegal. The biggest one of odd Integer is "301".      <p class="sycode">        </p>   <br>   <p class="sycode">    Input   </p>   <p class="sycode">    There are multiple test cases. Please process till EOF.    <br> Each case starts with a line containing an integer N ( 1 <= N <= 100 ).    <br /> The second line contains N Digits _1, a_2, a_3, \cdots, a_n. ( 0 \leqwhich indicate the digit $a  a_i \leq 9)$.   </p>   <p class="sycode">        </p>   <br />   <p class="sycode">    Output   </p>   <p class="sycode">    The output of each test case of a line. If you can constitute an Integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.   </p>   <p class="sycode">        </p>   <br />   <p class="sycode">    Sample Input   </p>   <p class="sycode">    <pre style="代码" class="precsshei">                  <p class="sycode">                      30 1 335 4 232 4 6                  </p>

                  <p class="sycode">                      301425-1                  </p>

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<vector>#include<cmath>#include<string>#include<queue>#define eps 1e-9#define ll long long#define INF 0x3f3f3f3fusing namespace std;const int maxn=100+10;int a[maxn],odd[maxn];char str[maxn];int n;int  main(){    int ans,pd;    while(scanf("%d",&n)!=EOF)    {        memset(str,0,sizeof(str));        int cnt=0,first=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if(a[i]%2)               odd[++cnt]=a[i];        }        sort(odd+1,odd+1+cnt);        sort(a+1,a+1+n);        int ly=n-1;        if(cnt==0)            puts("-1");        else        {            str[ly]=odd[1]+'0';            ly--;            for(int i=1;i<=n;i++)            {                if(a[i]==odd[1]&&!first)                {                    first=1;                    continue;                }                else                {                    str[ly]=a[i]+'0';                    ly--;                }            }            if(str[0]=='0')                puts("-1");            else                {                    for(int i=0;i<=n-1;i++)                        printf("%c",str[i]);                    printf("\n");                }        }    }    return 0;}

                  <p class="sycode">                      3abc1abcabc1abcabc2                  </p>

                  <p class="sycode">                      61521                  </p>