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    B. Pashmak and Flowers
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Pashmak decided to give Parmida a pair of flowers from the garden. There are n flowers in the garden and the i-th of them has a beauty number bi. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!

    Your task is to write a program which calculates two things:

    1. The maximum beauty difference of flowers that Pashmak can give to Parmida.
    2. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
    Input

    The first line of the input contains n (2 ≤ n ≤ 2·105). In the next line there are n space-separated integersb1, b2, ..., bn (1 ≤ bi ≤ 109).

    Output

    The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.

    Sample test(s)
    input
    2
    1 2
    output
    1 1
    input
    3
    1 4 5
    output
    4 1
    input
    5
    3 1 2 3 1
    output
    2 4
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<string>
    #include<cmath>
    #include<algorithm>
    #include<queue>
    #include<set>
    #include<vector>
    using namespace std;
    #define INF 0xfffffff
    #define LL long long
    LL n;
    LL a[200010];
    int main()
    {
         scanf("%d",&n);
         for(LL i=0;i<n;i++)
                scanf("%d",&a[i]);
         sort(a,a+n);
         if(a[0]==a[n-1])
         {
               printf("0 %I64d
    ",n*(n-1)/2);
         }
         else
         {
               LL maxx=0,minn=0,t1,t2;
               t1=0;
               while(t1<=n-1&&a[t1]==a[0])
                      minn++,t1++;
               t2=n-1;
               while(t2>=0&&a[t2]==a[n-1])
                      maxx++,t2--;
               printf("%I64d %I64d
    ",a[n-1]-a[0],minn*maxx);
         }
         return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/a972290869/p/4240750.html
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