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Mr. Wang is teaching simple arithmetic . Careful Teacher Wang collected i Taoist students often do wrong oral arithmetic , And I want to compile it into an exercise . It's tedious to arrange these topics , For this reason, he wants to use computer programs to improve work efficiency . Mr. Wang hopes to reduce the workload of input as much as possible , such as \texttt{5+8}5+8 You'd better just input \texttt 55 and \texttt 88, The output results should be as detailed as possible to facilitate the use of later typesetting , For example, the above input is processed and then output \texttt{5+8=13}5+8=13 And the total length of the formula 66. Mr. Wang gives you this glorious task , Please program for him to realize the above functions .

Input format

First line value ii

And then ii Actions require input formulas , Each row may have three data or two data .

If the behavior has three data, the first data represents the operation type ,\texttt aa For addition operations ,\texttt bb For subtraction ,\texttt cc For multiplication , The next two data represent the number of operands participating in the operation .

If the behavior has two data , The operation type of this question is the same as that of the previous question , And these two data are operands .

Output format

Output 2\times i2×i That's ok . For each input expression , Output complete formula and result , The second line outputs the total length of the expression

I/o sample

Input #1 Copy

Output #1 Copy

64+46=110
9
275+125=400
11
11*99=1089
10
46-64=-18
9

explain / Tips

Data scale and agreement

about 50\%50% The data of , The input formula has three data , The first formula must have three data .

For all the data ,0<i\leq 500<i≤50, The operands are non negative integers and less than 1000010000.

# include <iostream>
 using namespace std;
# include <string>
//char q[50];
string q;
string m;
//char m[50];
 int main()
 {
 	int n;
 	cin>>n;
 string p,k,z;
 	int d,c;

 	for(int i=0;i<n;i++)
 	{
 		cin>>p;
 		if(p=="a")
 		{
 			cin>>q>>m;
 				int d=stoi(q);
 					//cout<<" Here we are " ;
 				int c=stoi(m);
 			int e=c+d;
 			
 		string l=to_string(e);
 	//	cout<<l;
 				cout<<d<<"+"<<c<<"="<<d+c<<endl;
 				cout<<q.length()+m.length()+2+l.length()<<endl;
 					z=p;
		}
	else if(p=="b")
		{
				cin>>q>>m;
				int d=stoi(q);
 				int c=stoi(m);
 				string l=to_string(d-c);
 				cout<<d<<"-"<<c<<"="<<d-c<<endl;
 				cout<<q.length()+m.length()+2+l.length()<<endl;
				 	z=p;	
		}
		else if(p=="c")
		{
				cin>>q>>m;
				int d=stoi(q);
 				int c=stoi(m);
 			//	cout<<c*d; 
 				string l=to_string(d*c);
 		//	cout<<l;
			 	cout<<d<<"*"<<c<<"="<<d*c<<endl;
 				cout<<q.length()+m.length()+2+l.length()<<endl;	
 					z=p;
		}
 		else
 		{
 				if(z=="a")
 		{
 				cin>>m;
 				int d=stoi(p);
 				int c=stoi(m);
 		string l=to_string(c+d);
 			cout<<d<<"+"<<c<<"="<<d+c<<endl;
 			cout<<p.length()+m.length()+2+l.length()<<endl;
 				//	z=p;
		}
	else if(z=="b")
		{
			cin>>m;
				int d=stoi(p);
 				int c=stoi(m);
 				string l=to_string(d-c);
 				cout<<d<<"-"<<c<<"="<<d-c<<endl;
 				cout<<p.length()+m.length()+2+l.length()<<endl;
				 //	z=p;	
		}
		else if(z=="c")
		{
					cin>>m;
				int d=stoi(p);
 				int c=stoi(m);
 				string l=to_string(d*c);
 				cout<<d<<"*"<<c<<"="<<d*c<<endl;
 				cout<<p.length()+m.length()+2+l.length()<<endl;	
 				//	z=p;
		}
	
 		
		}
// 		cin>>a[i];
	
	 }
 //	cin>>a;
// 	int len=strlen(a);
// 	cout<<len;
// 	
 }

P1308 [NOIP2011 Popularization group ] Count the number of words

Title Description

General text editors have the function of finding words , This function can quickly locate the position of specific words in the article , Some can also count the number of times a specific word appears in the article .

Now? , Please program to realize this function , The specific requirement is ： Given a word , Please output the number of times it appears in a given article and the location of its first appearance . Be careful ： When matching words , Case insensitive , But it requires a perfect match , That is to say, a given word must be exactly the same as an independent word in the article regardless of case （ See example 1）, If a given word is only a part of a word in the article, it is not a match （ See example 2）.

Input format

common 22 That's ok .

The first 11 Acts as a string , There are only letters in it , For a given word ;

The first 22 Acts as a string , It can only contain letters and spaces , Represents a given article .

Output format

a line , If a given word is found in the text, output two integers , Two integers are separated by a space , They are the number of times the words appear in the article and the first place they appear （ That is, when it first appears in the article , The position of the first letter of a word in the article , Location slave 00 Start ）; If the word doesn't appear in the text , Then output an integer directly -1−1.

I/o sample

Input #1 Copy

To
to be or not to be is a question

Output #1 Copy

2 0

Input #2 Copy

to
Did the Ottoman Empire lose its power at that time

Output #2 Copy

-1

explain / Tips

Data range

1\leq1≤ The length of the first line of words \leq10≤10.

1\leq1≤ The length of the article \leq10^6≤106.

noip2011 Popularization Group No 2 topic

# include <bits/stdc++.h>
using namespace std;
int main()
{
string n;
//cin>>n;
getline(cin,n);
   string m;
    getline(cin,m);

//cin>>m;
// Lowercase all  
//transform(n.begin(),n.end(),n.begin(),::tolower);
//
//
//transform(m.begin(),m.end(),m.begin(),::tolower);
//int l=n.length();
 for (int i=0;i<n.length();++i){
        n[i]=tolower(n[i]);
    }
    for (int i=0;i<m.length();++i){
        m[i]=tolower(m[i]);
    }
m=' '+m+' ';
n=' '+n+' ';
if(m.find(n)==string::npos)
{
	cout<<-1;
 } 
else
{
	int alp=m.find(n);  // Record the position of the first occurrence 
	int blp=m.find(n),q=0;  // Used to find the number of occurrences later  

	while(blp!=string::npos) 
	{
		++q;
		//cout<<q<<endl;
		blp=m.find(n,blp+1);
	}
	cout<<q<<" "<<alp;
}






	
}

P1125 [NOIP2008 Improvement group ] Silly little monkey

Title Description

Stupid little monkey has a small vocabulary , So every time I do multiple choice questions, I have a headache . But he found a way , It has been proved that , It's a very good way to choose the right option ！

The specific description of this method is as follows ： hypothesis \text{maxn}maxn It's the number of letters that appear most frequently in a word ,\text{minn}minn It's the number of letters that appear the least in a word , If \text{maxn}-\text{minn}maxn−minn It's a prime number , So stupid little monkey thinks it's a Lucky Word, Such a word is probably the right answer .

Input format

A word , Only lowercase letters are possible , And the length is less than 100100.

Output format

There are two lines , The first line is a string , Suppose the input word is Lucky Word, Then output Lucky Word, Otherwise output No Answer;

The second line is an integer , If the input word is Lucky Word, Output \text{maxn}-\text{minn}maxn−minn Value , Otherwise output 00.

I/o sample

Input #1 Copy

error

Output #1 Copy

Lucky Word
2

Input #2 Copy

olympic

Output #2 Copy

No Answer
0

explain / Tips

【 I/o sample 1 explain 】

word error The letters that appear most in \texttt rr There is 33 Time , The letter with the least number of occurrences appeared 11 Time ,3-1=23−1=2,22 Prime number .

【 I/o sample 2 explain 】

word olympic The letters that appear most in \texttt ii There is 11 Time , The letter with the least number of occurrences appeared 11 Time ,1-1=01−1=0,00 Not prime .

（ The original problem has been corrected ）

noip2008 Improve the first question

# include <bits/stdc++.h>
using namespace std;
char s[105];
int a[27];
int main()
{
cin>>s;
int l=strlen(s);
int c;
	for(int i=0;i<l;i++)
	{
	c=s[i]-=97;
		a[c]++;
	}
	sort(a,a+26);
//for(int i=0;i<26;i++)
//{
//	cout<<a[i]<<"---"<<i<<"  ";
//}
//cout<<a[25];
int max=a[25];
int min;
		for(int i=0;i<26;i++)
		{
			if(a[i]!=0)
			{
		min=a[i];
		break;		
			}
		}
		int y=max-min;
int r;
if(y==1||y==0)
{
		cout<<"No Answer"<<endl<<"0";	
}
else
{
for(r=2;r<y;r++)
{
	if(y%r==0)
	{
		cout<<"No Answer"<<endl<<"0";
		break;
	}
}	
	
	if(r==y)
	{
		cout<<"Lucky Word"<<endl<<y;
	}
}

return 0;	
	
}

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