我是 C++ 的绝对初学者。字面上地。才过了一个星期。 今天我在写一个程序来测试需要多少次迭代才能使某个数字回文。 这是代码:

#include <iostream>
#include <string>
#include <algorithm>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers 1 to 1000
*/
using namespace std;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    int i=0;
    while (numsum.value.compare(numsumreverse.value) !=0)
    {
        reversenum=num;
        reversenum.reverse();
        numsum.value=to_string(stoll(num.value,0,10)+stoll(reversenum.value,0,10));
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
}

int main()
{
    number temp;
    int i;
    for (i=1; i<1001; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    return 0;
}

对于 195 以下的数字，它会顺利进行。但是，如果是 196，我会收到错误消息。 它说:

terminate called after throwing an instance of 'std::out_of_range' what(): stoll

我不知道该怎么办。我尝试从 196 开始，但错误仍然存​​在。任何帮助将不胜感激。 :)

更新:这次我尝试使用 ttmath 库来完成。但是啊!它再次停在 195，甚至不报告错误!我可能正在做一些愚蠢的事情。如有任何意见，我们将不胜感激。这是更新后的代码:

#include <iostream>
#include <string>
#include <algorithm>
#include <ttmath/ttmath.h>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers 1 to 1000
*/
using namespace std;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

template <typename NumTy>
string String(const NumTy& Num)
{
    stringstream StrStream;
    StrStream << Num;
    return (StrStream.str());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    ttmath::UInt<100> tempsum, numint, reversenumint;
    int i=0;
    while (numsum.value.compare(numsumreverse.value) !=0)
    {
        reversenum=num;
        reversenum.reverse();
        numint=num.value;
        reversenumint=reversenum.value;
        tempsum=numint+reversenumint;
        numsum.value=String<ttmath::UInt<100> >(tempsum);
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
}

int main()
{
    number temp;
    int i;
    for (i=196; i<1001; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    return 0;
}

更新:已解决。一些研究表明 196 可能是 Lychrel Number .在暗示 ttmath 库后我得到的结果只是让我确信我的算法有效。我已经尝试了所有高达 10000 的数字，它给出了完美的结果。这是最终代码:

#include <iostream>
#include <string>
#include <algorithm>
#include <ttmath/ttmath.h>
#include <limits>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers inside a desired range
*/
using namespace std;
string LychrelList;
int LychrelCount=0;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

template <typename NumTy>
string String(const NumTy& Num)
{
    stringstream StrStream;
    StrStream << Num;
    return (StrStream.str());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    ttmath::UInt<100> tempsum, numint, reversenumint;
    int i=0;
    while ((numsum.value.compare(numsumreverse.value) !=0) && i<200)
    {
        reversenum=num;
        reversenum.reverse();
        numint=num.value;
        reversenumint=reversenum.value;
        tempsum=numint+reversenumint;
        numsum.value=String<ttmath::UInt<100> >(tempsum);
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    if (i<200) cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
    else
    {
        cout << "A solution for " << n << " could not be found!!!" << endl;
        LychrelList=LychrelList+n+" ";
        LychrelCount++;
    }
}

int main()
{
    cout << "From where to start?" << endl << ">";
    int lbd,ubd;
    cin >> lbd;
    cout << endl << "And where to stop?" << endl <<">";
    cin >> ubd;
    cout << endl;
    number temp;
    int i;
    for (i=lbd; i<=ubd; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    if (LychrelList.compare("") !=0) cout << "The possible Lychrel numbers found in the range are:" << endl << LychrelList << endl << "Total - " << LychrelCount;
    cout << endl << endl << "Press ENTER to end the program...";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    string s;
    getline(cin,s);
    cout << "Thanks for using!";
    return 0;
}

这是一个非常棒的社区。特别鸣谢Marco A . :)

再次更新:我设计了自己的 add() 函数来减少程序对外部库的依赖。它导致更小的可执行文件和更快的性能。这是代码:

#include <iostream>
#include <string>
#include <algorithm>
#include <limits>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers inside a desired range
*/
using namespace std;
string LychrelList;
int LychrelCount=0;

string add(string sA, string sB)
{
    int iTemp=0;
    string sAns;
    int k=sA.length()-sB.length();
    int i;
    if (k>0){for (i=0;i<k;i++) {sB="0"+sB;}}
    if (k<0) {for (i=0;i<-k;i++) {sA="0"+sA;}}
    for (i=sA.length()-1;i>=0;i--)
    {
        iTemp+=sA[i]+sB[i]-96;
        if (iTemp>9)
        {
            sAns=to_string(iTemp%10)+sAns;
            iTemp/=10;
        }
        else
        {
            sAns=to_string(iTemp)+sAns;
            iTemp=0;
        }
    }
    if (iTemp>0) {sAns=to_string(iTemp)+sAns;}
    return sAns;
}

void palindrome(string num)
{
    string n=num;
    string reversenum, numsum, numsumreverse;
    numsum=num;
    numsumreverse=numsum;
    reverse(numsumreverse.begin(),numsumreverse.end());
    int i=0;
    while ((numsum.compare(numsumreverse) !=0) && i<200)
    {
        reversenum=num;
        reverse(reversenum.begin(),reversenum.end());
        numsum=add(num,reversenum);
        numsumreverse=numsum;
        reverse(numsumreverse.begin(),numsumreverse.end());
        num=numsum;
        i++;
    }
    if (i<200) cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num << endl;
    else
    {
        cout << "A solution for " << n << " could not be found!!!" << endl;
        LychrelList=LychrelList+n+" ";
        LychrelCount++;
    }
}

int main()
{
    cout << "From where to start?" << endl << ">";
    int lbd,ubd;
    cin >> lbd;
    cout << endl << "And where to stop?" << endl <<">";
    cin >> ubd;
    cout << endl;
    string temp;
    int i;
    for (i=lbd; i<=ubd; i++)
    {
        temp=to_string(i);
        palindrome(temp);
    }
    if (LychrelList.compare("") !=0) cout << "The possible Lychrel numbers found in the range are:" << endl << LychrelList << endl << "Total - " << LychrelCount;
    cout << endl << endl << "Press ENTER to end the program...";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    string s;
    getline(cin,s);
    cout <<endl << "Thanks for using!";
    return 0;
}

你们帮了我很多，让我找到了自己的路。感谢大家。 :)

最佳答案

自 num.value 和 reversenum.value 的最后两个有效值以来，您正在溢出 long long > 是 7197630720180367016 和 6107630810270367917，它们加在一起远远超过 long long 的最大大小(在我的机器上是 9223372036854775807)。这将产生负值并破坏您对 stoll

std::out_of_range is thrown if the converted value would fall out of the range of the result type or if the underlying function (std::strtol or std::strtoll) sets errno to ERANGE.

( reference )

• 如果你想得到下一个最小的回文，你应该使用另一种方法，比如I explained here。 .

你可以找到一个 Live Example 这里

• 如果您喜欢/必须继续您的方法，您应该在字符串上手动添加或使用 bigint 库(再次查看 here 并修改 plusOne() 功能根据你的喜好)

关于c++ - std::_throw_out_of_range 不知从何而来，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32807269/