2.20 learning content

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One 、 Negative modulo

In mathematics , The remainder of a negative number is a positive number . But in c++ in , Modulus of negative number is negative .
For example, in Mathematics ,-10 Divide 3 Remainder is 2; And in the c++ in ,-10%3=-1.

#include<iostream>
using namespace std;
int main()
{
    
	cout<< -10%3 <<endl;
	return 0;
}

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Two 、pow Function USES

Use pow Function time , The exponent position must be a floating point number .

#include<iostream>
#include<cmath>
using namespace std;
int main() 
{
    
	int x1 = 1;
	float x2 = 1.0;
	double x3 = 1.0;
	
	for(int i=1;i<=10;i++) 
		cout<<pow(i,x1/2)<<"\t"<<pow(i,x2/2)<<"\t"<<pow(i,x3/2)<<endl;
	
	return 0;
}

besides , I also read some other blogs , For bloggers printf Output directly with %d There will be a loss of accuracy , But there was a mess in my own compiler . So if you want to use printf The output still needs %f or %lf, Or add... Before it （int） It is safer to make a strong transition .

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3、 ... and 、 Some recently written questions

1.[NOIP2005] Trees outside the school gate

Topic link

The main idea of the topic ： There is a section of road where a tree is planted every other meter （ Including endpoint ）, The title will give you some intervals , Let you remove the trees in these intervals （ Including endpoint ）, These intervals may overlap , Ask how many trees are left .

Their thinking ： Set this path as an array and initialize it to 0, Let the points in each interval be ++, The final statistic is 0 The number of .
very violence convenient .

#include<iostream> 
#include<cstring>
using namespace std;

int main()
{
    
	int len,t,l,r;
	cin>>len>>t;
	int a[len+1];
	memset(a,0,sizeof(a));
	while(t--)
	{
    
		cin>>l>>r;
		for(int i=l;i<=r;i++) a[i]++;
	}
	
	int sum=0;
	for(int i=0;i<=len;i++)
		if(a[i]==0) sum++;
		
	cout<<sum<<endl;
	
	return 0;
}

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2. A simple big number addition problem

I know it's easy to add big numbers, but because I'm a good cook, I WA Don't laugh at me for asking for big chores many times

Topic link

General meaning ： There are two that are no longer than 5000 Add the numbers of , Both numbers may have leading zeros , Please output its sum , But do not output leading zeros .

Their thinking ： Input with a character array , Add bitwise and write in a int Type array , Operate on this array ： If it is greater than 10 Then carry . Finally, delete the leading zero .

#include<iostream>
#include<cstring>
using namespace std;
char x[5010],y[5010];
int ans[5010];

int main(){
    
    int t;
    cin>>t;
    while(t--)
	{
    
        cin>>x>>y;
        int len1=strlen(x);
		int len2=strlen(y);
        int i=len1-1,j=len2-1;
        
        int k=0;
        while(i>=0 && j>=0) ans[k++]+=x[i--]-'0'+y[j--]-'0';
        
        while(i>=0) ans[k++]+=x[i--]-'0';
        
        while(j>=0) ans[k++]+=y[j--]-'0';
        
        for(i=0;i<=k;i++)
		{
    
            if(ans[i]>=10)
			{
    
                ans[i+1] += ans[i]/10;// carry 
                ans[i] = ans[i]%10;
            }
        }
        
        while(ans[k]==0) k--;
        // It is very convenient to delete leading zeros 
        for(i=k;i>=0;i--)cout<<ans[i];
        cout<<endl;
        
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        memset(ans,0,sizeof(ans));
    }
    return 0;
}

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