問(wèn)題描述

我想弄清楚如何使用遞歸來(lái)執(zhí)行 n 級(jí)嵌套 for 循環(huán).例如，如果 n=3，就會(huì)有 3 個(gè)級(jí)別"

I'm trying to figure out how I can use recursion to do n-level nested for loops. For example, if n=3, there would be 3 'levels'

for(z=0;z<6;z++){
   for(y=0;y<6;y++){
      for(x=0;x<6;x++){
         if (z+y+x==f){
            //do something
         } 
      }
   }
}

等等.

我似乎無(wú)法弄清楚如何將 if 循環(huán)放置在最后一個(gè) for 循環(huán)中，以及如何從 if 語(yǔ)句訪問(wèn)前一個(gè) for 循環(huán)的變量.我知道變量嵌套循環(huán)的問(wèn)題已經(jīng)被問(wèn)過(guò)很多次了，我已經(jīng)瀏覽了所有這些問(wèn)題.但似乎沒(méi)有人幫助我.

I can't seem to figure out how I would be able to place the if loop in the last for loop and how I can access the variables of previous for loops from the if statement. I know that the question of variable nested loops has been asked alot of times, and I have looked through all of them. But none seem to help me.

有人可以提出一種使用遞歸來(lái)實(shí)現(xiàn)這一目標(biāo)的簡(jiǎn)單方法，請(qǐng)記住我仍然是 C++ 的初學(xué)者，為我指明正確的方向嗎?

Could someone present an easy way of using recursion to achieve this, keeping in mind that I'm still a beginner in c++, to point me in the right direction?

用例如下:

編寫程序輸入骰子的個(gè)數(shù)m.程序?qū)⑤敵隹赡馨咐目倲?shù)、每個(gè)可能 n 的可能案例數(shù)量以及具有最高概率的 n.注意:只有一個(gè)輸入 m 被讀入. n 由程序計(jì)算

Write a program to input the number of dice m. The program will output the total number of possible cases, the number of possible cases for each possible n and the n with the highest probability. Note: only one input m is read in. n is computed by the program

示例如果用戶輸入 m=2 則程序應(yīng)輸出

Example if user enters m=2 then program should output

可能的病例總數(shù)為 36.
可能性是
2 1
3 2
4 3
.
.
.
12 1

The total number of possible cases is 36.
The possibilities are
2 1
3 2
4 3
.
.
.
12 1

推薦答案

這是一個(gè)普通的舊 C++ 示例.首先，我為每個(gè)維度創(chuàng)建一個(gè)范圍向量，稱為 maxes.如果所有索引的總和為 2，那么我打印做了一些事情.在示例中，我將 z 從 0 循環(huán)到 1，將 y 從 0 循環(huán)到 2，將 x 從 0 循環(huán)到 3

Here's an example in plain old C++. First I make a vector of the ranges for each dimension called maxes. if the sum of all indices are 2 then I print did something. In the example I loop z from 0 to 1, y from 0 to 2, x from 0 to 3

你肯定可以把它弄得更整潔.

You can for sure make this more neat.

這里是:

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

int f(){ 
    return 2 ;
}

void inner(int depth,vector<int> & numbers,vector<int> & maxes){
  if (depth>0){
     for(int i=0;i<maxes[depth-1];i++){
        numbers[depth-1]=i;
        inner(depth-1, numbers,maxes) ;
     }
  }else{
     // calculate sum of x,y,z:
     cout << "values are ";
     for(int i=0;i<numbers.size();i++){
        cout <<numbers[i]<<" ";
     }
     int thesum(0);
     for(int i=0;i<numbers.size();i++){
        thesum+=numbers[i];
     }
     if (thesum==f()){
        cout << "did something! ";
     }
     cout<<endl;
   }
}

void donest(){
   vector<int>  numbers;
   numbers.resize(3);
   vector<int>  maxes;
   maxes.push_back(4);
   maxes.push_back(3);
   maxes.push_back(2);
   inner(numbers.size(),numbers,maxes);
}

int main(){
   donest();
}

結(jié)果:

values are 0 0 0 
values are 1 0 0 
values are 2 0 0  did something! 
values are 3 0 0 
values are 0 1 0 
values are 1 1 0  did something! 
values are 2 1 0 
values are 3 1 0 
values are 0 2 0  did something! 
values are 1 2 0 
values are 2 2 0 
values are 3 2 0 
values are 0 0 1 
values are 1 0 1  did something! 
values are 2 0 1 
values are 3 0 1 
values are 0 1 1  did something! 
values are 1 1 1 
values are 2 1 1 
values are 3 1 1 
values are 0 2 1 
values are 1 2 1 
values are 2 2 1 
values are 3 2 1 

這篇關(guān)于變量嵌套 for 循環(huán)的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！