zoukankan      html  css  js  c++  java
  • Enemy at the Gateway

    Enemy at the Gateway
    Time Limit:8000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

    Description

    Download as PDF

    Problem H
    Enemy at the Gateway 
    Input: 
    Standard Input

    Output: Standard Output

    This is the year 9002, the war between the Earth and Triton has broken out. Being a spy for the Earth, you are trying to find all means to destroy Triton protections. Being a very smart spy, you have somehow managed to enter into the Triton network and take over the control of a gateway. Now, you are trying to decode the messages passed between them.

    o

    The messages are arbitrarily long sequence of integers with a preamble of P integers. Each integer will fit inside a 32 bit signed integer. You have managed to capture the actual preamble. But, the communication lines are too noisy, and for this reason, you can not get the sequence accurately. For each number in the sequence si, you have determined that, it can actually be any value between pi and qi inclusive.

    Now, given the sequence of numbers, find in how many places, the message may start.

    Input

    First line contains T, the number of test cases. Each test case starts with an integer P, the length of the pattern. The next line contains Pintegers. Next line contains 10 integers, Np0q0ABCDEFMN  is the length of the sequence. The range for each element in the sequence is generated using a generator function

    pi = (A * pi-1 + B * qi-1 + C) % M

    qi = (D * pi-1 + E * qi-1 + F) % M

    if(qi < pi) swap(pi,qi)

    [p1,q1], [p2, q2],...,[pN,qN] describes the sequence. Please note that [p0,q0] is not included in the sequence

    Output

    For each test case, output the number of places, the preamble may start.

    Constraints

    l      T <= 100

    l      P <= 60

    l      N <= 1000000

    l      0 <= p0,q0,A,B,C,D,E,F,M <= 1000000000

    Sample Input                              Output for Sample Input

    2

    1

    10

    4 6 12 1 0 2 0 1 2 100000

    2

    1 2

    3 1 3 1 0 0 0 1 0 10

    Case 1: 2

    Case 2: 2

    In the first case. the intervals are:

    [8,14],[10,16],[12,18],[14,20]. The value 10 can only be contained in the first two.

    In the second case, all the intervals are [1,3], so, you can find two positions to start the preamble.


    Problem setter: Manzurur Rahman Khan

    Special Thanks: Samee Zahur

    数据水了,暴力。。。

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    int T, t = 1;
    int N, A, B, C, D, E, F, M, p;
    int a[61];
    int ans;
    
    //区间
    struct Range
    {
       int p, q;
    }r[1000001];
    
    int main()
    {
        int i, j;
        scanf("%d", &T);
        while(T--)
        {
            ans = 0;
            scanf("%d", &p);
            for(i = 0; i < p; i++)
                scanf("%d", &a[i]);
            scanf("%d %d %d %d %d %d %d %d %d %d", &N, &r[0].p, &r[0].q, &A, &B, &C, &D, &E, &F, &M);
            for(i = 0; i < N; i++)
            {
                r[i+1].p = ( ( (A * r[i].p) % M + (B * r[i].q) % M) % M + C) % M;
                r[i+1].q = ( ( (D * r[i].p) % M + (E * r[i].q) % M) % M + F) % M;
                if(r[i+1].p > r[i+1].q)  swap(r[i+1].p, r[i+1].q);
            }
            //注意loop的条件是i <= N - p + 1
            for(i = 1; i <= N - p + 1; i++)
            {
                int k = i;
                for(j = 0; j < p; j++, k++)
                {
                    if(a[j] > r[k].q || a[j] < r[k].p)
                        break;
                }
                if(j == p) ans++;
            }
            printf("Case %d: %d\n", t++, ans);
        }
        return 0;
    }
    


  • 相关阅读:
    lombok-@Accessors注解
    spring boot 当参数传入开头多个0时,报错:JSON parse error: Invalid numeric value: Leading zeroes not allowed
    linux查看历史操作记录并且显示执行时间
    IDEA中mybatis插件自动生成手写sql的xml文件
    CPU核数和load average的关系
    Jenkins--Credentials添加证书从git上拉代码
    解决输入git branch 进入编辑状态,mac下出现END,无法返回
    Git log和git reflog
    SpringCloud入门之常用的配置文件 application.yml和 bootstrap.yml区别
    springboot定时任务
  • 原文地址:https://www.cnblogs.com/cszlg/p/2910484.html
Copyright © 2011-2022 走看看