C++ C/C和x2B中的Euler#11项目+；
我有这段代码是为了解决ProjectEuler的第11个问题，它的工作范围仅限于C++ C/C和x2B中的Euler#11项目+；,c++,algorithm,C++,Algorithm,我有这段代码是为了解决ProjectEuler的第11个问题，它的工作范围仅限于printout（），之后它会因“堆栈溢出”而崩溃。 我猜我的索引（可能在next_elem函数中）有问题，但我看不出有什么问题。请给我一个提示 #include "stdafx.h" #include <iostream> const int matrix[20][20] = { {8, 2, 22, 97, 38, 15, 0, 40,
printout（）next_elem#include "stdafx.h"
#include <iostream>
const int matrix[20][20] = {
{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
{52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
{24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
{67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
{24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
{21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
{78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
{16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
{86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
{19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
{4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
{88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
{4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
{20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
{1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48},
};
unsigned long int rows[20][20], columns [20][20], diagslash[20][20], diagbackslash[20][20]; //create all necessary arrays of maximum possible dimensions
//a debugging function -- to check that everything is right and that I'm working with indices correctly
void printout()
{
for(int i=0;i<20;++i)
{
{
for(int j=0;j<20;++j)
std::cout << "a["<<i<<"],["<<j<<"] = " << matrix[i][j] << " " << "\n";
}
std::cout << "\n";
}
}
// a function that calculates the next element from the previous one and some matrix elements
unsigned long int next_elem(int i,int j, unsigned long int array[][20])
{
if (matrix[i][j] == 0) // a zero in a sequence means we need to move 4 elements to the right
{
j+=4;
next_elem(i,j,rows);
}
else if(matrix[i][j] > matrix[i][j-4]) // this condition is meant to put an increasing sequence of results into the 'rows' array
{
rows[i][j-4]=rows[i][j-5]*matrix[i][j-1]/matrix[i][j-1];
next_elem(i,j,rows);
}
return 0;
}
void check(int i, int j, unsigned long int array[][20])
{
if ((matrix[i][j] == 0) || (matrix[i][j+1] == 0) || (matrix[i][j+2] == 0) || (matrix[i][j+3] == 0)) //look for a zero
{
j+=4;
check(i,j,rows);
}
else //no zero found, calculate the next element
{
rows[i][j] = matrix[i][j]*matrix[i][j+1]*matrix[i][j+2]*matrix[i][j+3];
next_elem(i,j+4,rows);
}
}
void sort() //this function is meant to find maximums in 'rows', 'columns' and so on
{
for (int i = 0; i<20;++i)
{
for (int j = 0;j < 17;++j)
{
check(i,j,rows);
std::cout << "rows["<<i<<"]["<<j << "] = " << rows[i][j] << std::endl; //a debugging line -- to see what's going on
}
}
}
int main()
{
printout();
sort();
return 0;
}
#包括“stdafx.h”
#包括
常数整型矩阵[20][20]={
{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
{52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
{24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
{67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
{24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
{21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
{78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
{16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
{86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
{19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
{4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
{88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
{4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
{20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
{1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48},
};
无符号长整型行[20][20]，列[20][20]，对角斜杠[20][20]，对角反斜杠[20][20]//创建所有必要的最大可能维度数组
//一个调试功能——检查一切是否正常，以及是否正确使用索引
无效打印输出（）
{
对于（int i=0；i，很难跟踪所有条件，但很明显，您可以很容易地越界：
void check(int i, int j, unsigned long int array[][20])
{
if ((matrix[i][j] == 0) || (matrix[i][j+1] == 0) || (matrix[i][j+2] == 0) || (matrix[i][j+3] == 0)) //look for a zero
{
j+=4; // what if j was 16 here?
check(i,j,rows);
}
}
因此，作为第一步，我建议使用assert检查范围内的所有参数：
void check(int i, int j, unsigned long int array[][20])
{
assert( i >= 0 && i < 20 && j >= 0 && j < 20 );
...
}
void检查（int i、int j、无符号长int数组[][20]）
{
断言（i>=0&&i<20&&j>=0&&j<20）；
...
}
对下一个\u元素执行相同的操作
在启用调试的情况下编译并检查。在这种情况下，不会对i、j或矩阵执行任何操作：
else if(matrix[i][j] > matrix[i][j-4]) // this condition is meant to put an increasing sequence of results into the 'rows' array
{
rows[i][j-4]=rows[i][j-5]*matrix[i][j-1]/matrix[i][j-1];
next_elem(i,j,rows);
}
所以它将永远递归，总是使用相同的参数，总是通过相同的分支，最终给你一个堆栈溢出；这将是我的猜测
您需要逐步执行i
或j
，或者更改矩阵[i][j]
或矩阵[i][j-4]
，这样条件最终将为false，从而终止递归
没有进一步分析代码或问题，因此我不知道您需要更改什么：）您是seg故障，因为：
for (int j = 0;j < 17;++j)
for（int j=0；j<17；++j）
需要
for (int j = 0; j < 16; ++j)
for（int j=0；j<16；++j）
//m是矩阵
int max=0；
无效更新（int i）{
如果（i>max）max=i；}
void main（）{
对于（int i=0；我总是发布完整的错误消息。在MS VS 2012中，本地调试器显示：0xc0000fd:Stack overflow（参数：0x00000001、0x00252FE4）你能澄清一下吗：你说你的程序在调用打印输出时崩溃了，但是你怀疑罪魁祸首可能是下一个元素
函数-但是你没有在打印输出
内部调用下一个元素
，而打印输出
是你首先要调用的l insidemain
噢，天哪。对不起，必须纠正这个问题。您正在调用一个函数check（）
递归的，所以堆栈溢出在这里总是一个风险，如果它没有正确实现。使用调试程序现在我很困惑：我认为外部循环是为了改变I
和j
@Chiffa；什么外部循环？它是一个递归，你用相同的参数从内部再次调用next\elem
，没有循环Involved@Chiffa；如果您指的是sort
中的循环，恕我冒犯，但我相信您有点过头了，您需要后退一步，了解参数是如何传递的，递归是什么。@Chiffa；即使sort
中的循环会更改next\elem
中的值（他们没有这样做，只是使用新值调用next\u elem
，将循环值传播到next\u elem
），你没有给循环一个步进的机会，你在自己调用next\u elem
，它永远不会返回让循环运行。@Chiffa；我真的不知道算法应该如何工作，但你需要将新值传递给next\u elem
，你不能给它相同的值。在next\elem中更改j
将产生这种效果ct，但你似乎认为变量j
与t有某种联系
// m is matrix
int max=0;
void update(int i){
if(i>max)max=i;}
void main(){
for(int i=0;i<17;i++){
for(int j=0;j<17;j++){
update(m[i][j]*m[i+1][j]*m[i+2][j]*m[i+3][j]);
update(m[i][j]*m[i][j+2]*m[i][j+3]*m[i][j+4]);
update(m[i][j]*m[i+1][j+1]*m[i+2][j+2]*m[i+3][j+3]);
update(m[i][j+3]*m[i+1][j+2]*m[i+2][j+1]*m[i+3][j]);
}
}
printf("%d\n",max);
}