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  • 2014 ACM-ICPC Beijing Invitational Programming Contest

    点击打开链接

    Happy Reversal

    1000ms
    65536KB
    64-bit integer IO format: %lld      Java class name: Main
    Font Size:  
    Type:  
    Elfness is studying in an operation "NOT".
    For a binary number A, if we do operation "NOT A", after that, all digits of A will be reversed. (e.g. A=1001101, after operation "NOT A", A will be 0110010).
    Now Elfness has N binary numbers of length K, now he can do operations "NOT" for some of his numbers. 
    Let's assume after his operations, the maximum number is M, the minimum number is P. He wants to know what's the maximum M - P he can get. Can you help him?

     

    Input

    The first line of input is an integer T (T ≤ 60), indicating the number of cases.
    For each case, the first line contains 2 integers N (1 ≤ N ≤ 10000) and K (1 ≤ K ≤ 60), the next N lines contains N binary numbers, one number per line, indicating the numbers that Elfness has. The length of each binary number is K.
     

    Output

    For each case, first output the case number as "Case #x: ", and x is the case number. Then you should output an integer, indicating the maximum result that Elfness can get.

    Sample Input

    2
    5 6
    100100
    001100
    010001
    010001
    111111
    5 7
    0001101
    0001011
    0010011
    0111000
    1001011
    

    Sample Output

    Case #1: 51
    Case #2: 103
    

    Source

    给你n组由k个0或者1组成的二进制数。每一个数能够翻转。求两个数的最大差值。

    //152 ms 1788 KB
    #include<stdio.h>
    #include<algorithm>
    using namespace std;
    char s[107];
    long long ans[20007];
    int n,m;
    long long getnum()
    {
        long long res=0,j=1;
        for(int i=m-1;i>=0;i--,j*=2)
            if(s[i]=='1')res+=j;
        return res;
    }
    int main()
    {
        int t,cas=1;
        scanf("%d",&t);
        while(t--)
        {
            int k=0;
            scanf("%d%d",&n,&m);
            for(int i=0;i<n;i++)
            {
                scanf("%s",s);
                ans[k++]=getnum();
                for(int j=0;j<m;j++)
                    if(s[j]=='1')s[j]='0';
                    else s[j]='1';
                ans[k++]=getnum();
            }
            sort(ans,ans+k);
            long long ans1=ans[k-1]-ans[1];
            long long ans2=ans[k-2]-ans[0];
            long long p=ans[0],q=ans[k-1];
            long long e=1<<(m-1);
            if((p^q)&e==e)
                printf("Case #%d: %lld
    ",cas++,max(ans1,ans2));
            else printf("Case #%d: %lld
    ",q-p);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/ldxsuanfa/p/10029238.html
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