Try to solve PTA 20211122 - Basic Practice of function, which involves high-precision addition and conversion from decimal system to 2-hexadecimal system. Students in need can have a look;

catalogue

1, Foreword

2, Example part

2.1 binary conversion

2.1.2 problem solving 2

2.2,   Function returns the inverse of an integer (20 minutes)

2.2.2   Problem solution

2.3,   Large integer A+B (10 points)

2.3.1 problem solving

two point four   Find the number of combinations (16 points)

2.4.1   Problem solution

3, Epilogue

1, Foreword

Hello, I'm Xia MI. Call me "Xia" or "Xiao Mi".

This time, we will introduce the routine PTA test every week.

Let's look at today's five questions.

2, Example part

2.1 binary conversion

Converts A decimal integer n (− 231 ≤ n ≤ 231 − 1) to A k (2 ≤ k ≤ 16) hexadecimal number. Note that 10 ~ 15 are represented by letters A, B, C, D, E and F respectively.

Input format:

First, enter a positive integer T to represent the number of groups of test data, and then T groups of test data. Enter two integers n and k for each set of test data.

Output format:

For each group of tests, first output n, then output a space, and finally output the corresponding k-ary number.

```4
5 3
123 16
0 5
-12 2
```

```5 12
123 7B
0 0
-12 -1100```

2.1.1 problem solving 1

Since 11 ~ hexadecimal conversion involves A, B, C, D, E and F letters, we need to use string to solve this problem:

First, we compile a universal binary conversion function, which is of type string.

We use t=n% hexadecimal number to obtain the last digit T, such as 3 for decimal 123;

n / = hexadecimal number    To destroy the last digit, such as decimal 123, destroy 3, and become 12;

Small tips: t is int type. To convert it to char type, you can do this

t+‘0’   For example, t=1, 1 + '0' = '1';

ans+= '1'；；

If t > 10, t-10 + 'A'; For example, 10 - 10 + 'A' = 'A';

ans+= 'A'；

```string intToA(int n,int radix)
{
string ans="";	//Define an empty string ans;
do{
int t=n%radix;	//Take the last digit of the decimal system, such as 3 in 123 of the decimal system;
if(t>=0&&t<=9)    ans+=t+'0';//If the last digit of the base is 0 ~ 9, add ans directly; T is int type, t+'0' is transformed into char type;
else ans+=t-10+'A';//If the last digit of the hexadecimal is greater than 9, it is converted to the corresponding hexadecimal representation, such as 10 to A,11 to B, etc;
n/=radix;//Destroy the last bit, e.g. 123 destroy 3 to 12;
}while(n!=0);//Use do while to ensure that ans="0" is returned when the input is 0;
reverse(ans.begin(),ans.end());//The subfunction of the string class reverses the elements of the string, such as 321 to 123;
return ans; //Returns ans;
}```

Then, it is specially judged that the input value is less than 0:

Just add a '-' sign before the absolute value result;

```#include<bits/stdc++.h>
using namespace std;
{
string ans="";	//Define an empty string ans;
do{
int t=n%radix;	//Take the last digit of the decimal system, such as 3 in 123 of the decimal system;
if(t>=0&&t<=9)    ans+=t+'0';//If the last digit of the base is 0 ~ 9, add ans directly; T is int type, t+'0' is transformed into char type;
else ans+=t-10+'A';//If the last digit of the hexadecimal is greater than 9, it is converted to the corresponding hexadecimal representation, such as 10 to A,11 to B, etc;
n/=radix;//Destroy the last bit, e.g. 123 destroy 3 to 12;
}while(n!=0);//Use do while to ensure that ans="0" is returned when the input is 0;
reverse(ans.begin(),ans.end());//The subfunction of the string class reverses the elements of the string, such as 321 to 123;
return ans; //Returns ans;
}
int main(){
int n=0;
scanf("%d",&n);
for(int i=0;i<n;i++){
int num,k;
scanf("%d %d",&num,&k);
if(num>=0)cout<<num<<" "<<intToA(num,k)<<endl;
else cout<<num<<" -"<<intToA(num*-1,k)<<endl;
}
}```

2.1.2 problem solving 2

Look at this function!

It is included in this header file #include < stdlib. H >

itoa(num,s,k)；// Num is the integer int or long, s is the character array, and K is the hexadecimal number to be converted;

The return type is string;

```#include<bits/stdc++.h>
using namespace std;
int main(){
int num,k;
char s[100];
cin>>num>>k;
cout<<itoa(num,s,k);//num is the integer int or long, s is the character array, and k is the hexadecimal number to be converted
}```

However, when it outputs A number such as base 11, the output 11 is' A 'instead of' A ';

So we write a small loop to replace case;

```#include<bits/stdc++.h>
using namespace std;
{
string ans="";
do{
if(t>=0&&t<=9)    ans+=t+'0';
else ans+=t-10+'a';
}while(n!=0);
reverse(ans.begin(),ans.end());
return ans;
}
int main(){
int n=0;
scanf("%d",&n);
for(int i=0;i<n;i++){
int num,k;
char s[999];
scanf("%d %d",&num,&k);

switch(k){
case 2:{
itoa(num,s,2);
printf("%s",s);
break;
}
case 3:{
itoa(num,s,3);
printf("%s",s);
break;
}
case 4:{
itoa(num,s,4);
printf("%s",s);
break;
}
case 5:{
itoa(num,s,5);
printf("%s",s);
break;
}

case 6:{
itoa(num,s,6);
printf("%s",s);
break;
}
case 7:{
itoa(num,s,7);
printf("%s",s);
break;
}
case 8:{
itoa(num,s,8);
printf("%s",s);
break;
}
case 9:{
itoa(num,s,9);
printf("%s",s);
break;
}

case 10:{
itoa(num,s,10);
printf("%s",s);
break;
}
case 11:{
itoa(num,s,11);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
case 12:{
itoa(num,s,12);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
case 13:{
itoa(num,s,13);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
case 14:{
itoa(num,s,14);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
case 15:{
itoa(num,s,15);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
case 16:{
itoa(num,s,16);
for(int i=0;s[i];i++){
if(s[i]>='a'&&s[i]<='z')s[i]+=('A'-'a');
}
printf("%s",s);
break;
}
}
}
}```

2.2,   Function returns the inverse of an integer (20 minutes)

Write a function that returns the inverse of a formal parameter (a positive integer). The main function is to input an integer N and output the inverse number of N.

Input example:

`21000`

No blank lines at the end

Output example:

`12`

No blank lines at the end

Input example:

`1234`

No blank lines at the end

Output example:

`4321`

No blank lines at the end

2.2.2   Problem solution

```#include<bits/stdc++.h>
using namespace std;
int rev(int x){
int a[1000],i=0,sum=0;
while(x!=0)
{
a[i]=x%10;
i++;
x/=10;
}
for(int j=0;j<i;j++){
sum+=a[j];
sum*=10;
}
sum/=10;
return sum;
}

int main(){
int n;
cin>>n;
cout<<rev(n);
}```

Refer to the solution in 2.1. Note that because we record the array in reverse order, we can write it in this method

```for(int j=0;j<i;j++){
sum+=a[j];
sum*=10;
}
sum/=10;
return sum;```

Come and see vector

```#include<bits/stdc++.h>
using namespace std;
int main(){
long long int sum=0,n;
vector<int> a;
cin>>n;
while(n!=0){
a.push_back(n%10);
n/=10;
}
for(auto x:a){
sum+=x;
sum*=10;
}
cout<<sum/10;
} ```

After that, I will talk about vector for interested students;

2.3,   Large integer A+B (10 points)

Enter two integers A and B to find A + B.

Input format:

First, enter A positive integer T to represent the number of groups of test data, and then T groups of test data. Enter 2 positive integers A and B for each test group. Integers may be large, but the number of bits per integer will not exceed 1000.

Output format:

Output two lines of data for each group of tests; The first line outputs "Case #:", # represents the test group number, the second line outputs "A+B = Sum", and Sum represents the result of A+B. There is a blank line between each two sets of test data.

Input example:

```2
1 2
88888888888888888888888 11111111111111111111111
```

Output example:

```Case 1:
1 + 2 = 3

Case 2:
88888888888888888888888 + 11111111111111111111111 = 99999999999999999999999
```

HDOJ 1002

2.3.1 problem solving

```#include<bits/stdc++.h>
using namespace std;
string P  (string & num,string add){
int g=0;
string t=num;
}//If the length of num is less than the length of add, exchange the contents of num and add;
add= t+add;//Complete the length of add in the digital head to make it as long as num; If 123 is completed as 000123;
for(int i=len1-1;i>=0;i--){
num[i]=t%10+'0';//Calculate from the low position;
g=t/10;//Whether the record base is empty;
}
if(g!=0){
num.insert(0,string(1,(char)g+'0'));
}//Judge the system of the head;
return num;
}
int main(){
int n,n1=1,p=0;
cin>>n;
for(int i=0;i<n;i++){
string a,b,c;
cin>>a>>b;
c=a;
if(p!=0)cout<<endl<<endl;
p=1;
printf("Case %d:\n",n1);
n1++;
cout<<c<<" + "<<b<<" = "<<P(a,b);
}
} ```

You can use the same idea to write high-precision subtraction, high-precision multiplication and high-precision division;

help!

two point four   Find the number of combinations (16 points)

This problem requires writing a program according to the formula Cnm = m!(n−m)!n! Calculate the combination number of M elements (m ≤ n) taken from n different elements.

It is recommended to define and call the function fact(n) to calculate n!, Where the type of n is int and the function type is double.

Input format:

The input gives two positive integers m and n (m ≤ n) on one line, separated by spaces.

Output format:

Output according to the format "result = calculation result of combination number". The title ensures that the result is within the range of double type.

Input example:

`2 7`

No blank lines at the end

Output example:

`result = 21`

No blank lines at the end

2.4.1   Problem solution

```#include<bits/stdc++.h>
using namespace std;
double fact(int n){
if(n>=1)return fact(n-1)*n;
else return 1;
}
int main(){
int n,m;
double sum=0;
cin>>m>>n;
sum=(fact(n))/((fact(m))*fact(n-m));
printf("result = %.lf",sum);
}```

Water problem, let's play by ourselves!