2014 ACM-ICPC Beijing Invitational Programming Contest

最新推荐文章于 2025-08-08 22:01:10 发布

weixin_30859423

最新推荐文章于 2025-08-08 22:01:10 发布

阅读量190

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/ldxsuanfa/p/10029238.html

本文探讨了一种算法问题，即通过翻转一组由0和1组成的二进制数，寻找两个数之间的最大差值。文章提供了一个具体的实现方案，包括输入输出格式、样例和源代码，展示了如何计算并返回最大可能的差值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击打开链接

Happy Reversal

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

Prev Submit Status Statistics Discuss Next

Font Size: + -

Type: NoneGraph Theory 2-SAT Articulation/Bridge/Biconnected Component Cycles/Topological Sorting/Strongly Connected Component Shortest Path Bellman Ford Dijkstra/Floyd Warshall Euler Trail/Circuit Heavy-Light Decomposition Minimum Spanning Tree Stable Marriage Problem Trees Directed Minimum Spanning Tree Flow/Matching Graph Matching Bipartite Matching Hopcroft–Karp Bipartite Matching Weighted Bipartite Matching/Hungarian Algorithm Flow Max Flow/Min Cut Min Cost Max Flow DFS-like Backtracking with Pruning/Branch and Bound Basic Recursion IDA* Search Parsing/Grammar Breadth First Search/Depth First Search Advanced Search Techniques Binary Search/Bisection Ternary Search Geometry Basic Geometry Computational Geometry Convex Hull Pick's TheoremGame Theory Green Hackenbush/Colon Principle/Fusion Principle Nim Sprague-Grundy Number Matrix Gaussian Elimination Matrix Exponentiation Data Structures Basic Data Structures Binary Indexed Tree Binary Search Tree Hashing Orthogonal Range Search Range Minimum Query/Lowest Common Ancestor Segment Tree/Interval Tree Trie Tree Sorting Disjoint Set String Aho Corasick Knuth-Morris-Pratt Suffix Array/Suffix Tree Math Basic Math Big Integer Arithmetic Number Theory Chinese Remainder Theorem Extended Euclid Inclusion/Exclusion Modular Arithmetic Combinatorics Group Theory/Burnside's lemma Counting Probability/Expected Value Others Tricky Hardest Unusual Brute Force Implementation Constructive Algorithms Two Pointer Bitmask Beginner Discrete Logarithm/Shank's Baby-step Giant-step Algorithm Greedy Divide and Conquer Dynamic Programming Tag it!

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

Sample Output

Case #1: 51
Case #2: 103

Source

2014 ACM-ICPC Beijing Invitational Programming Contest

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\n",cas++,max(ans1,ans2));
        else printf("Case #%d: %lld\n",q-p);
    }
    return 0;
}

转载于:https://www.cnblogs.com/ldxsuanfa/p/10029238.html