我正在研究欧拉项目Problem 14,我需要找到1,000,000下最长的collatz序列。我想出了一种适用于较小数字(比如100)的算法,该算法将每个1到100之间的collatz数存储到一个数组中,并将该数组用作参考,以加快计算数量。 我的代码如下:分割错误C++(数组太大?)
#include <iostream>
using namespace std;
long even(long n){ //the even-collatz function
n=n/2;
return n;
}
long odd(long n){ //the odd collatz function
n=3*n+1;
return n;
}
int main(){
long x, c=0, y[1000000]; // x= the number we are finding the collatz number of, c a counter that keeps track of how many steps we've taken in the sequence, y is an array to store the collatz numbers.
for (x=1; x<1000000; x++){ //iterates from x=1 to 1 million
long a = x; //sets a number a equal to the number we are currently trying to find the collatz number of
long b = a;
c=0; //intializes counter at 0
while (a!=0){ //loops infinitely; the only way to exit is through a break.
if (a%2==0){ // detects if the number is even
a=even(a); //applies the even-collatz function if so; sets x=x/2
c=c+1;
if (y[a]!=0){ // checks if the collatz number of x is already discovered
y[b]=c+y[a]; //adds the current number of steps to the collatz number of x and
break; //exits the while loop
}
}
else if (a==1){ //checks if the new x is equal to one and
y[b]=c; //if it is, it writes the current value of c to y[b] and
break; // exits the loop
}
else if (a%2==1){ //same as the "even" block, except for odd numbers
a=odd(a);
c=c+1;
if(y[a]!=0){
y[b]=c+y[a];
break;
}
}
//this is the end of the while loop; we've applied the collatz function as many times as we've needed to to x, and incremented the counter each time
}
}
long z;
for (int n=0;n!=100;n++){
if (y[n+1]>y[n]){
z=y[n+1];
}
}
cout << z << "\n";
}
我遇到的问题是,我得到一个segfault在x = 1818在for循环。通过调试,我发现段错误发生的速度取决于数组y的大小,所以我假设该数组太大了。从我对segfaults的基本理解来看,我认为我只是访问了我“不被允许”的内存。有什么方法可以绕过这个问题吗?或者我应该开始努力寻找解决这个问题的另一种解决方案吗?我正在Ubuntu Studio上使用g ++进行编译。
堆栈溢出导致许多平台上的分段错误。 – 2014-09-28 00:33:38
@HevyLight是的,但它不依赖于x有多大 - 只要函数运行,您就会得到异常。如果分段错误是由缓冲区溢出导致的,而不是堆栈溢出。 – 2014-09-28 02:00:28