C++ Program Basics (17): C++ Loop Nesting

Definition of loop nesting

Literally speaking, nesting refers to the method of embedding one object into another object when one or more tables, images or layers are added to an existing table, image or layer, or when two objects have an assembly relationship.

"Loop nesting" can also be understood in this way, that is, embedding one or even multiple loops in one loop.

In this article, onlyfor loop nestingAs an example.

Classic double for loop

First, let’s take a look at a story:

The weather has been hot recently, and Xiao T is sweating profusely every day. Today, Xiao T saw his ownA great injusticeThe best friend Xiao E passed by,RobberyAsk Xiao E to buy a popsicle for himself. After a little thought, Xiao E said: “Although I don't have money,I can print one for you! "Little T tenangrySaid happily: "I want one$4\ast 4$Popsicles of size! ”

You can help Xiao E print a for Xiao T$4\ast 4$A popsicle of size?

OK, no matter whether you want it or notNow let’s analyze this question:

According to the question, we need to print a rectangle with four rows and four columns (output * here). It is obviously not enough to rely solely on a single-layer cycle, so it requires a classic double cycle to play.

for(int i=1;i<=4;i++){
	for(int j=1;j<=4;j++){
			cout<<"*";//Inner layer loop control column, that is, the * output from each row
	}
	cout<<endl;//The outer loop control line can make the inner loop 4 times
}

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AI writing code

As shown in the above code:

The outer loop controls the number of rows, so that the inner loop executes$4$times; the inner layer loop controls the number of columns, that is, the output of each row*number.

Multiple for loops

Let's take a look at another story:

Xiao S was born into a very large and wealthy family, and next week will be the day for their family to hold a big family banquet. The young talented Xiao S has taken on the family affairs since childhood. For the upcoming family dinner, Xiao S needs to use limited use.$100$Buy one hundred chickens for yuan as ingredients. Among them, three little chickens are sold for one yuan, one hen is sold for three yuan, and one rooster is sold for five yuan. How many of the three chickens should be bought?

In order to help your young master Xiao S buy all the ingredients, let’s analyze the question now:

In this question, using double cycles is not enough, we need to use triple cycles (representing the number of roosters, hens and chicks respectively):

for(int i=0;i<=100/5;i++){//The number of roosters purchased
	for(int j=0;j<=100/3;i++){//The number of hens purchased
		for(int k=0;k<=300;k++){//The number of chickens purchased
			if(i+j+k==100&&i*5+j*3+k/3==100&&k%3==0){
					cout<<i<<" "<<j<<" "<<k<<endl;
			}
		}
	}
}

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The outermost cycle first determines the number of roosters to buy, and then the inner layer is performed with two cycles to find out whether there are hens and chicks that can meet the conditions (the innermost cycle isif judgmentWhether it meets the question conditions)

If it cannot meet, increase the number of roosters purchased and continue the cycle until a logarithm is found.

example

When Xiao U saw the "popsicle" that Xiao T made for Xiao E, he was envious and asked Xiao T if he could make one for himself, and Xiao T agreed happily. But little U said that He wants one$4\ast 4$A triangle of size "popsicles". Now Xiao T is in trouble. Can you help him again?

Output sample

*
**
***
****

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Question explanation
We can set up a double loop: the outer loop$i$Depend on$1$to$4$Incremental, the inner layer of$j$Then$1$to$i$Increment and output simultaneously*；

for(int i=1;i<=a;i++){
        for(int j=1;j<=i;j++){
            cout<<"*";
        }
        cout<<endl;
    }

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This will achieve the effect of outputting triangles.