Central Europe Regional Contest 2011-白红宇

Central Europe Regional Contest 2011

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发布时间：2019-06-28

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Unique Encryption Keys

Time Limit: 30000ms, Special Time Limit:75000ms, Memory Limit:65536KB

Total submit users: 19, Accepted users: 15

Problem 12484 : Special judge

Problem description

The security of many ciphers strongly depends on the fact that the keys are unique and never re-used. This may be vitally important, since a relatively strong cipher may be broken if the same key is used to encrypt several different messages.

In this problem, we will try to detect repeating (duplicate) usage of keys. Given a sequence of keys used to encrypt messages, your task is to determine what keys have been used repeatedly in some specified period.

Input

The input contains several cipher descriptions. Each description starts with one line containing two integer numbers M and Q separated by a space. M (1≤M≤1000000) is the number of encrypted messages, Q is the number of queries (0≤Q≤1000000).

Each of the following M lines contains one number Ki (0≤i≤2³⁰) specifying the identifier of a key used to encrypt the i-th message. The next Q lines then contain one query each. Each query is specified by two integer numbers Bj and Ej, 1≤Bj≤Ej≤M, giving the interval of messages we want to check.

There is one empty line after each description. The input is terminated by a line containing two zeros in place of the numbers M and Q.

Output

For each query, print one line of output. The line should contain the string ``OK" if all keys used to encrypt messages between B_j and E_j (inclusive) are mutually different (that means, they have different identifiers). If some of the keys have been used repeatedly, print one identifier of anysuch key.

Print one empty line after each cipher description.

Sample Input

10 532349738411 32 64 103 72 65 2123122 41 50 0

Sample Output

3OK43OKOK1

Problem Source

Central Europe 2011

题意：告诉你从1-n的区间内，依次对应一个整数，询问给定一个区间，是否存在两个相同的整数。

1 #include 
     
       2 #include 
      
        3 #include 
        4 #include 
        
          5 #include 
         
           6 using namespace std; 7  8 int n,m; 9 vector
          
            value;10 11 int main()12 {13 for(;;) 14 {15 scanf("%d %d\n", &n, &m);16 if (!n) return 0;17 value.resize(n);18 for (int i = 0; i < n; i++) 19 {20 scanf("%d", &value[i]);21 }22 23 map
           
             last_occ; //用于记录当前的最小匹配区间24 vector
            
              first_match(n); //用于记录从i位置之后最先匹配的区间的右侧25 for (int i = n-1; i >= 0; i--) 26 {27 first_match[i] = n;28 if (i < n-1)29 first_match[i] = first_match[i+1];30 if (last_occ.count(value[i]))31 first_match[i] = min(first_match[i], last_occ[value[i]]); //更新区间32 last_occ[value[i]] = i; 33 }34 35 for (int i = 0; i < m; i++)36 {37 int x, y;38 scanf("%d %d",&x, &y);39 x--;40 y--;41 if (first_match[x] <= y)42 {43 printf("%d\n",value[first_match[x]]);44 } 45 else46 {47 printf("OK\n");48 }49 }50 printf("\n");51 }52 }53 /*54 12 1055 256 657 1258 559 660 461 762 663 764 1465 266 1467 3 768 */

题目链接：

注意一下vector和map的用法

1、vector<int>value(n); value大小为n 等同于 vector<int>value; value.resize(n);

2、map<int,int>cnt; 等同于建立一个映射（(int)x->(int)y）

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