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Finishing (III) - Exercise 2
2022-07-04 06:55:00 【Dan·】
1、 Enter some positive integers (≤26), Program to output the letter triangle attached to the wall with these positive integers as the side length .
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n,i,j;
while(cin>>n)
{
char a;
for(i=1;i<=n;i++)
{
cout<<"a";
for(j=1;j<=n-i;j++)
{cout<<" ";}
char a=97;
for(j=1;j<=i;j++)
{cout<<char(a+j-1);}
cout<<endl;
}
cout<<endl;
}
return 0;
}
Input :
3
7
Output :
a a
a ab
aabca a
a ab
a abc
a abcd
a abcde
a abcdef
aabcdefg
2、 Xiao Jin plays cards , When the number of cards in the other party's hand is greater than or equal to 10 Zhang Shi , He will observe each other's expression , If the other party looks nervous , He's out “ Fried ”, If the other party is laughing , He doesn't play any cards .
character 'J' Express that the other party is nervous about applying ,'H' Express that the other party is laughing . Xiao Jinchu “ Fried ”, The output “Z”, If you don't play any cards , The output “pass”.
When the number of cards in the other hand is less than 10 Zhang's time , He will “ Pick up ” The other side's card , That is, the other party goes out “ single ”, He also came out “ single ”, The other party “ double ”, He also made a double .
character 'D’ On behalf of the other party “ single ”,'S' On behalf of the other party “ double ”. Xiao Jinchu “ single ”, The output “D”, If it comes out “ double ”, The output “S”.
#include<iostream>
using namespace std;
int main()
{
int n;char b,c;
cin>>n>>b>>c;
if(n>=10)
{
if(b=='J') cout<<"Z";
if(b=='H') cout<<"pass";
}
if(n<10)
{
if(c=='D') cout<<"D";
if(c=='S') cout<<"S";
}
}
Input 1:
15 H D
Input 2:
9 J S
Output 1:
pass
Output 2:
S
3、 A hundred dollars for a hundred chickens : A chicken is worth five , A hen is worth three , A chicken is worth three , A hundred dollars for a hundred chickens , Ask jiweng 、 mother 、 How much is it ?
#include<iostream>
using namespace std;
int main()
{
for (int a = 0; a <= 20; a++)
for (int b = 0; b <= 33; b++)
for (int c = 0; c <= 100 - a - b; c = c + 3)
if ((a + b + c == 100) and (5 * a + 3 * b + c / 3 == 100))
cout << a << " "<<b<<" " << c;
}
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