為什么我的代碼首先執行后面的cout？

Question

以下代碼

#include "stdafx.h"
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>

using namespace std;

bool checkPerm(unsigned long long x){
    vector<unsigned long long> tester;
    string strx = to_string(x);
    int sizestrx = strx.size();
    int counter = 1;
    cout << "x is " << strx << " and its permutations are ";
    while (next_permutation(strx.begin(), strx.end())){
        cout << strx << " ";
        unsigned long long stoipermstrx = stoi(strx);
        tester.push_back(stoipermstrx);
    }
    cout << endl;
    int sizetester = tester.size();
    for (int j = 2; j <= 6; j++){
        cout << "j is " << j << ' ';
        for (int k = 0; k < sizetester; k++){
            if (j*x == tester[k]){
                cout << "counter increased because x, counter " << x << " " << counter << endl;
                counter++;
                if (counter == 6){
                    cout << "Number is " << x << endl;
                    return true;
                }
                break;
            }
        }
        //cout << "Number " << x << " failed" << endl;
        return false;
    }
    return true;
}

int main(){
    unsigned long long x = 1;
    for (double i = 0; ; i++){
        cout << i << endl;
        while (x < 1.67*pow(10, i)){
            if (i == 5)
                cout << x << endl;
            if (checkPerm(x)){
                cin.get();
            }
            x++;
        }
        x = pow(10, (i + 1));
    }
    cin.get();
}

在這段代碼中存在以下問題：

cout << "x is " << strx << " and its permutations are ";
while (next_permutation(strx.begin(), strx.end())){
    cout << strx << " ";
    unsigned long long stoipermstrx = stoi(strx);
    tester.push_back(stoipermstrx);
}
cout << endl;
int sizetester = tester.size();
for (int j = 2; j <= 6; j++){
    cout << "j is " << j << ' ';
    for (int k = 0; k < sizetester; k++){
        if (j*x == tester[k]){
            cout << "counter increased because x, counter " << x << " " << counter << endl;
            counter++;
            if (counter == 6){
                cout << "Number is " << x << endl;
                return true;
            }
            break;
        }
    }
    //cout << "Number " << x << " failed" << endl;
    return false;
}

這里的輸出將是“ j是jx是x，其排列是（x的排列）”。 但是，控制台應打印“ x為x，其排列為（排列）j為j”。 給出以下示例輸出：

j is 2 x is 1355 and its permutations are 1535 1553 3155 3515 3551 5135 5153 531
5 5351 5513 5531
j is 2 x is 1356 and its permutations are 1365 1536 1563 1635 1653 3156 3165 351
6 3561 3615 3651 5136 5163 5316 5361 5613

Answer 1

看來這有兩件事。 第一，在打印j的值之前您沒有查看sizetester的值，並且在j的值之后也不打印換行符。 這意味着您將在當前'x'的行首顯示上一個循環的j值。 如果我了解您的代碼應該執行的操作，則似乎正確地執行了它-這只是顯示輸出的方式而使它感到困惑。

嘗試這個：

int sizetester = tester.size();
for (int j = 2; j <= 6; j++){
    if (sizetester){                           // <-- added test (see below)
        cout << "j is " << j << '\n';          // <-- added newline
    }                                          // <--

針對sizetester的測試可抑制j值的虛假打印-以后無論如何(k < sizetester)測試(k < sizetester) 。 換行符只是防止j 的值在x的下一個值處開始 ，這似乎是導致輸出混亂的原因。