1. Programming questions (5 questions in total, 100 points in total)
1. Digital amplification
Given a sequence of integers and an amplification factor x, amplify each integer in the sequence x times and output it.
Time limit: 1000
Memory limit: 65536
[Input]
contains three lines: The first line is N, indicating the length of the integer sequence (N ≤ 100); the second line is N integers (not exceeding the integer range), and the integers are separated by one. Separated by spaces; the third line contains an integer (not exceeding the integer range), which is the specified integer x.
[Output]
N integers, which are the amplified sequences of the original sequence. The integers are separated by a space.
[Sample input]
3
1 5 7
2
[Sample output]
2 10 14
提示： 注意答案的数据范围
Reference answer
#include<bits/stdc++.h>
using namespace std;
long long a[105];
int main(){
long long n , x;
cin >> n;
for(int i = 0;i < n;i++){
cin >> a[i];
}
cin >> x;
for(int i = 0;i < n;i++){
cout << a[i] * x << " ";
}
return 0;
}
2. Word inversion
Write a program to read a line of English (containing only letters and spaces, with words separated by single spaces), reverse the order of all words and output them, still separated by single spaces.
Time limit: 10000
Memory limit: 65536
[Input]
Input is a string (string length is at most 100).
[Output]
The output is a string sorted as required.
[Sample input]
I am a student
[Sample output]
student a am I
Reference answer
#include<bits/stdc++.h>
using namespace std;
string s[100005];
int main(){
int n = 0;
while(cin >> s[n]){
n++;
}
for(int i = n - 1;i >= 0;i--){
cout << s[i] << " ";
}
return 0;
}
3. Calculate the sum of edge elements of the matrix
The so-called elements at the edge of the matrix are the elements in the first and last rows and the elements in the first and last columns.
Time limit: 10000
Memory limit: 65536
[Input]
The first line is the row number m and column number n of the matrix respectively (m < 100, n < 100), separated by a space. In the next m lines of data input, each line contains n integers, separated by a space.
【Output】
Output the sum of edge elements of the corresponding matrix
[Sample input]
3 3
3 4 1
3 7 1
2 0 1
[Sample output]
15
Reference answer
#include<bits/stdc++.h>
using namespace std;
int a[105][105];
int main(){
int n , m , sum = 0;
cin >> n >> m;
for(int i = 0;i < n;i++){
for(int j=0;j<m;j++){
cin >> a[i][j];
}
}
for(int i = 0;i < n;i++){
for(int j = 0;j < m;j++){
if(i == 0 || i == n - 1 || j == 0 || j == m - 1){
sum = sum + a[i][j];
}
}
}
cout << sum;
return 0;
}
4. Odd single increasing sequence
Given a sequence of positive integers of length N (not greater than 500), please take out all the odd numbers and output them in ascending order.
Time limit: 1000
Memory limit: 65536
[Input]
There are 2 lines in total: The first line is N; the second line is N positive integers, separated by spaces.
【Output】
A sequence of odd numbers output in increasing order, separated by commas. The data is guaranteed to have at least one odd number.
[Sample input]
10
1 3 2 6 5 4 9 8 7 10
[Sample output]
1,3,5,7,9
Reference answer
#include<bits/stdc++.h>
using namespace std;
int a[505];
int main(){
int n;
int ans = 0;
cin >> n;
for(int i = 0;i < n;i++){
cin >> a[i];
}
sort(a + 0 , a + n);
for(int i = 0;i < n;i++){
if(a[i] % 2 != 0){
if(ans == 1){
cout<<",";
}
cout << a[i];
ans = 1;
}
}
return 0;
}
5. Addition of real numbers
Find the sum of two real numbers added together.
The floating point numbers appearing in the input and output in the question have the following forms: P1P2…Pi.Q1Q2…Qj. For the integer part, P1P2...Pi is a non-negative integer and when the integer part is not 0, P1 is not equal to 0; for the decimal part, Qj is not equal to 0.
Time limit: 1000
Memory limit: 65536
[Input]
2 lines, each line is an addend. The length of each addend cannot exceed 100.
【Output】
One row, the corresponding sum. The output is guaranteed to be a real number with a decimal part that is not 0.
[Sample input]
0.111111111111111111111111111111
0.111111111111111111111111111111
[Sample output]
0.222222222222222222222222222222
Reference answer
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
char a[202],b[202],c[101],d[101];
cin>>a;
cin>>b;
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
int i,j=0,ap=0,bp=0,ka=0,kb=0;
//ka,kb用来统计数组a和b的小数部分的位数
//ap，bp用来统计数组a和b的整数部分的位数
//统计数组a的整数部分和小数部分的位数
for(i=0; i<strlen(a); i++)
{
if(a[i]=='.')
{
break;
}
ap++;
}
ka=strlen(a)-ap-1;
//统计数组b的整数部分和小数部分的位数
for(i=0; i<strlen(b); i++)
{
if(b[i]=='.')
{
break;
}
bp++;
}
kb=strlen(b)-bp-1;
//对于整数部分从个位开始加即倒叙输入
for(i=0; i<ap||i<bp; i++)
{
if(i<ap)
{
c[i]+=a[ap-1-i]-'0';
}
if(i<bp)
{
c[i]+=b[bp-1-i]-'0';
}
if(c[i]>9) //大于9则向前进1
{
c[i+1]+=1;
c[i]%=10;
}
}
//判断是否有小数部分
if(ka>0||kb>0)
{
for(i=0; i<ka||i<kb; i++)
{
if(i<ka)
{
d[i]+=a[i+ap+1]-'0';//小数部分正序输入
}
if(i<kb)
{
d[i]+=b[i+bp+1]-'0';
}
}
//取两者最大小数位,用于输出小数输出
if(ka<kb)
{
ka=kb;
}
//取两者最大整数位
if(ap<bp)
{
ap=bp;
}
for(i=ka-1; i>0; i--) //向前进1
{
if(d[i]>9)//小数部分的进位
{
d[i-1]+=1;
d[i]%=10;
}
}
if(d[0]>9)//对第一位小数是否进位进行判断
{
c[0]+=1;
d[0]%=10;
while(c[j]>9)
{
c[j+1]+=1;
c[j]%=10;
}
}
if(c[ap]>0)
{
printf("%d",c[ap]);
}
for(i=ap-1; i>=0; i--)
{
printf("%d",c[i]);
}
printf(".");
for(i=0; i<ka; i++)
{
printf("%d",d[i]);
}
}
else
{
if(ap<bp)
{
ap=bp;
}
if(c[ap]>0)
{
printf("%d",c[ap]);
}
for(i=ap-1; i>=0; i--)
{
printf("%d",c[i]);
}
}
return 0;
}