zoukankan      html  css  js  c++  java
  • 赛码"BestCoder"杯中国大学生程序设计冠军赛1001——Movie

    Problem Description

    Cloud and Miceren like watching movies.

    Today, they want to choose some wonderful scenes from a movie. A movie has N scenes can be chosen, and each scene is associate with an interval [LR]. L is the beginning time of the scene and R is the ending time. However, they can't choose two scenes which have overlapping intervals. (For example, scene with [1, 2] and scene with [2, 3], scene with [2, 5] and scene with[3, 4]).

    Now, can you tell them if they can choose such three scenes that any pair of them do not overlap?

    Since there are so many scenes that you can't get them in time, we will give you seven parameters N, L1, R1, a, b, c, d, and you can generate L1 ~ LNR1 ~ RN by these parameters.

    Input

    The first line contains a single integer T, indicating the number of test cases.

    Each test case contains seven integers N, L1, R1, a, b, c, d, meaning that there are N scenes. The i-th scene's interval is [Li, Ri]. L1 and R1 have been stated in input, and Li = (Li1  a + b) mod 4294967296, Ri = (Ri1  c + d) mod 4294967296.

    After all the intervals are generated, swap the i-th interval's Li and Ri if Li > Ri.

    T is about 100.

    1  N  10000000.

    1  L1,R1  2000000000.

    1  a,b,c,d  1000000000.

    The ratio of test cases with N > 100 is less than 5%.

    Output

    For each test, print one line.

    If they can choose such three scenes, output "YES", otherwise output "NO".

    Sample Input
    2
    3 1 4 1 1 1 1
    3 1 4 4 1 4 1
    Sample Output
    NO
    YES
    大意:大水。。问是否存在三个区间满足不重叠
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int MAX = 10000000+10;
    const int inf = 0x3f3f3f3f;
    struct edge{
        unsigned int l,r;
    }t[MAX];
    int main()
    {
        int T,n;
        unsigned int a, b, c, d;
        scanf("%d",&T);
        while(T--){
            scanf("%d%d%d%d%d%d%d",&n,&t[1].l,&t[1].r,&a,&b,&c,&d);
            for(int i = 2; i <= n ; i++){
                t[i].l = t[i-1].l*a + b;
                t[i].r = t[i-1].r*c + d;
            }
            for(int i = 1; i <= n ;i++){
                if(t[i].l > t[i].r)
                    swap(t[i].l,t[i].r);
            }
            unsigned int min1 = 4294967295UL,max1 = 0;
            for(int i = 1; i <= n ; i++){
                if(t[i].r < min1)
                    min1 = t[i].r;
                if(t[i].l > max1)
                    max1 = t[i].l;
            }
            if(min1 > max1){
                printf("NO
    ");
                continue;
            }
            int flag = 0;
            for(int i = 1; i <= n ; i++){
                if(t[i].l > min1 && t[i].r < max1){
                    printf("YES
    ");
                    flag = 1;
                    continue;
                }
            }
            if(!flag) 
                printf("NO
    ");
        } 
    return 0;
    }
    
    
    

      

    
    
  • 相关阅读:
    vs2013运行qt.exe文件提示错误“找不到Qt5Cored.dll”
    基于stm32f1的lora开发基础通信实验
    vs出现找不到dll,重新安装程序可能会解决此问题
    C&C++&opencv文件操作
    出现无法解析的符号xxx(vs2013)
    opencv目标跟踪检测(C)
    vs运行出错:error MSB8020或error LNK1104: 无法打开文件“opencv_calib3d248d.lib/opencv_contribxxxd.lib”
    vs串口读写dll封装C++#(免费源码分享)
    stm32定时器操作配置()
    pwm控制ili9341背光屏幕亮度(stm32f4)
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4477222.html
Copyright © 2011-2022 走看看