2022.02.15

Brush questions as usual

Problem description

　　 one day ,JOE Finally can't stand the calculation a^b%c This ordinary operation . So he decided to ask you to write a program , Calculation a^b%c.

　　 Tips ： if b It's odd ,,a^b=(a^(b/2))^2*a, otherwise a^b=(a^(b/2))^2.

Input format

　　 Three nonnegative integers a,b,c;

Output format

　　 An integer ans, Express a^b%c;

The sample input

7 2 5

Sample output

4

Data size and engagement

　　30% a <= 100, b <= 10^4, 1 <= c <= 100
　　60% a <=10^4, b <= 10^5, 1 <= c <= 10^4
　　100% a <=10^6, b <= 10^9, 1 <= c <= 10^6

This problem involves fast exponentiation , In order to reduce the amount of calculation , The base number needs to be modeled constantly ; At the same time, we should also judge the parity of the index

#include<bits/stdc++.h>
using namespace std;

int main()
{
    long long a,b,c;
    cin>>a>>b>>c;
    long long sum=1;
    while(b)
    {
        if(b%2==0)
            a=(a*a)%c;
        else if(b%2!=0)
        {
            sum=(a*sum)%c;
            a=(a*a)%c;
        }
        b=b/2;
    }
    cout<<sum%c;
    return 0;
}

  Error summary ： The first time I want to write a function , The fact proved that , Don't write recursion when you have nothing to do , Otherwise the computer will burn

                    The second time it exploded , Facts have proved that long long why int

Problem description

　　 Enter any 10 A floating point number , According to their degree of aggregation, they are divided into 3 Group , Output the average value of each group .
　　 Provide a train of thought for teachers to talk in class ： take 10 Number to sort on the number axis , Then calculate the distance between every two points , Disconnect at the two largest distances of all distances , This way 10 The numbers are divided into 3 Group .
　　 This question is difficult , If the in-depth discussion is complicated , You can only consider the very simple case of grouping as shown in the following example , As long as the simple case can be calculated successfully , You can score on this question .

　　 in addition , This topic is a little ahead of schedule , It is recommended that you learn about one-dimensional arrays in the first section of the chapter on arrays by yourself , Then use a one-dimensional array to do . Sorting algorithm can refer to trustie Bubble sorting method reference uploaded by the platform .

Input format

　　 Ten floating-point numbers to be input , Use spaces to separate

Output format

　　 The average of the three groups , A new line is required for each output

The sample input

An example of input that meets the requirements of the topic .
example 1：
50.4 51.3 52.3 9.5 10.4 11.6 19.1 20.8 21.9 49.6

example 2：
8.6 7.4 3.5 17.9 19.1 18.5 37.6 40.4 38.5 40.0

Sample output

The output corresponding to the sample input above .
example 1：
10.5
20.6
50.9

example 2:
6.5
18.5
39.125

Jicao

#include<bits/stdc++.h>
using namespace std;

int main()
{
    double a[10];
    int i;
    for(i=0;i<10;i++)
        cin>>a[i];
    sort(a,a+10);
    double b[10],max1=0,max2=0,sum=0;
    for(i=0;i<9;i++)
        b[i]=fabs(a[i]-a[i+1]);
    int x1,x2,temp;
    x1=x2=0;
    for(i=0;i<9;i++)
    {
        if(max2<b[i]&&max1>b[i])
        {
            max2=b[i];
            x2=i;
        }
        if(max1<b[i])
        {
            max2=max1;
            x2=x1;
            max1=b[i];
            x1=i;
        }
    }
    if(x1>x2)
    {
        temp=x1;
        x1=x2;
        x2=temp;
    }              //x2>x1

    for(i=0;i<=x1;i++)
        sum+=a[i];
    cout<<sum/(x1+1)<<endl;
    sum=0;
    for(i=x1+1;i<=x2;i++)
        sum+=a[i];
    cout<<sum/(x2-x1)<<endl;
    sum=0;
    for(i=x2+1;i<10;i++)
        sum+=a[i];
        cout<<sum/(9-x2);
    return 0;
}

The main points of ： Judge the partition interval