PTA question bank== > complex four operations, one for one, examination seat number (7-73)

One 、 Four operations of complex number

1. Title Description

This question requires the preparation of procedures , Calculation 2 The sum of two plurals 、 Bad 、 product 、 merchant .

2. Input format

Type in a line and press a1 b1 a2 b2 The format of 2 Plural C1=a1+b1i and C2=a2+b2i The real part and the virtual part of . Title assurance C2 Not for 0.

3. Output format

Respectively in 4 In line with (a1+b1i) Operator (a2+b2i) = result Format order output 2 The sum of two plurals 、 Bad 、 product 、 merchant , The number is accurate to the decimal point 1 position . If the real or imaginary part of the result is 0, No output . If the result is 0, The output 0.0.

4. Examples

sample input 1

2 3.08 -2.04 5.06

  sample output 1

(2.0+3.1i) + (-2.0+5.1i) = 8.1i
(2.0+3.1i) - (-2.0+5.1i) = 4.0-2.0i
(2.0+3.1i) * (-2.0+5.1i) = -19.7+3.8i
(2.0+3.1i) / (-2.0+5.1i) = 0.4-0.6i

sample input 2

1 1 -1 -1.01

sample output 2

(1.0+1.0i) + (-1.0-1.0i) = 0.0
(1.0+1.0i) - (-1.0-1.0i) = 2.0+2.0i
(1.0+1.0i) * (-1.0-1.0i) = -2.0i
(1.0+1.0i) / (-1.0-1.0i) = -1.0

5. Code section

#include <stdio.h>
#include <stdlib.h>

void Print_same(double a1, double b1, double a2, double b2, char c);

void Print_result(double res1, double res2);

double rounding(double num);

int main()
{
	double a1, b1, a2, b2;
	scanf("%lf%lf%lf%lf", &a1, &b1, &a2, &b2);
	
	Print_same(a1, b1, a2, b2, '+');
	Print_result(rounding(a1 + a2), rounding(b1 + b2));
	 
	Print_same(a1, b1, a2, b2, '-');
	Print_result(rounding(a1 - a2), rounding(b1 - b2));
	
	Print_same(a1, b1, a2, b2, '*');
	Print_result(rounding(a1 * a2 - b1 * b2), rounding(a1 * b2 + a2 * b1));
	
	Print_same(a1, b1, a2, b2, '/');
	Print_result(rounding((a1 * a2 + b1 * b2) / (a2 * a2 + b2 * b2)), rounding((-a1 * b2 + a2 * b1) / (a2 * a2 + b2 * b2)));
	return 0;
}


void Print_same(double a1, double b1, double a2, double b2, char c) {
	if (b1 < 0 && b2 < 0) printf("(%.1lf%.1lfi) %c (%.1lf%.1lfi) = ", a1, b1, c, a2, b2);
	else if (b1 < 0) printf("(%.1lf%.1lfi) %c (%.1lf+%.1lfi) = ", a1, b1, c, a2, b2);
	else if (b2 < 0) printf("(%.1lf+%.1lfi) %c (%.1lf%.1lfi) = ", a1, b1, c, a2, b2);
	else printf("(%.1lf+%.1lfi) %c (%.1lf+%.1lfi) = ", a1, b1, c, a2, b2);
}
void Print_result(double res1, double res2) {
	if (res1 == 0 && res2 == 0) printf("0.0\n");
	else if (res1 == 0) printf("%.1lfi\n", res2);
	else if (res2 == 0) printf("%.1lf\n", res1);
	else if (res2 < 0) printf("%.1lf%.1lfi\n", res1, res2);
	else printf("%.1lf+%.1lfi\n", res1, res2);
}
double rounding(double num)
{
	if (num > 0) num = (int)(num * 10 + 0.5) / 10.0;
	else num = (int)(num * 10 - 0.5) / 10.0;
	return num;
}

 

Two 、 One gang, one

1. Title Description

“ A group of students ” It is a common way of learning organization in primary and secondary schools , The teacher put the students who are in the first place and the students who are in the second place . Please write a program to help the teacher finish the assignment automatically , That is, after getting the ranking of the whole class , Among the students who are not currently grouped , Compare the top students with the bottom students opposite sex The students are divided into groups .

2. Input format

Enter the first line to give a positive even number N（≤50）, The number of students in the class . thereafter N That's ok , Give the gender of each student in the order of ranking from high to low （0 On behalf of girls ,1 On behalf of the boy ） And name （ No more than 8 A non empty string of English letters ）, In the meantime 1 Space separation . It is guaranteed that the ratio of men to women in this class is 1:1, And there's no parallel .

3. Output format

Output the names of two students in each line , In the meantime 1 Space separation . The high ranking students are in front , The students with lower rank are in the back . The output order of the group is from the top to the bottom according to the ranking of the students in front .

4. Examples

sample input

8
0 Amy
1 Tom
1 Bill
0 Cindy
0 Maya
1 John
1 Jack
0 Linda

sample output

Amy Jack
Tom Linda
Bill Maya
Cindy John

#include <stdio.h>

struct Student
{
    int sex;
    char name[9];
};

int main()
{
    int n, i, j;
    scanf("%d", &n);
    struct Student student[n];
    for (i = 0; i < n; i++)
    {
        scanf("%d %s", &student[i].sex, student[i].name);
    }
    for (i = 0; i < n; i++)
    {
        for(j = n - 1; j >= i; j--)
        {
            if (student[i].sex != student[j].sex && student[j].sex != 3)
            {
                printf("%s %s\n", student[i].name, student[j].name);
                student[j].sex = 3;
                break;
            }
        }
    }
    return 0;
}

 

3、 ... and 、 Test seat number

1. Title Description

Every PAT Candidates are assigned two seat numbers when they take the exam , One is a test seat , One is the exam seat . Under normal circumstances , Candidates will get the seat number of the test machine before entering , After entering the test state , The system will display the test seat number of the candidate , Candidates need to change their seats during the examination . But some candidates are late , The test run is over , They can only ask you for help with the number of the test seat they have received , Find out their test seat number from the background .

2. Input format

The first line of input gives a positive integer  N（≤1000）, And then  N  That's ok , Each line gives a candidate's information ： Ticket number Test seat number Test seat number . among Ticket number from 16 Digit composition , Seat from 1 To  N  Number . Input to make sure that everyone's admission number is different , And at no time will two people be assigned to the same seat .

After candidate information , Give a positive integer  M（≤N）, In the next line  M  Test seat numbers to be inquired , Space off .

3. Output format

Corresponding to each seat number to be inquired , Output the candidate's pass number and seat number in one line , Intermediate use 1 Space separation .

4. Examples

Input

4
3310120150912233 2 4
3310120150912119 4 1
3310120150912126 1 3
3310120150912002 3 2
2
3 4

Output

3310120150912002 2
3310120150912119 1

5. Code demonstration

#include<stdio.h>
struct Number
{
    long a;
    int b;
    int c;
};

int main()
{
    int n;
    scanf("%d", &n);
    struct Number number[n];
    for (int i = 0; i < n; i++) {
        scanf("%ld %d %d", &number[i].a, &number[i].b, &number[i].c);
    }
    int late;
    scanf("%d", &late);
    int arr2[late];
    for (int i = 0; i < late; i++) {
        scanf("%d", &arr2[i]);
    }

    for (int i = 0; i < late; i++) {
        for (int j = 0; j < n; j++) {
            if (arr2[i] == number[j].b) {
                printf("%ld %d\n", number[j].a, number[j].c);
            }
        }
    }
    return 0;
}