项目Euler任务#8,代码在某一点

时间:2016-02-17 18:29:30

标签: c++ debugging overflow codeblocks unsigned-long-long-int

我在任务中遇到某种问题  https://projecteuler.net/problem=8,(从1000个数字字符串中找到13个连续数字的最高乘积),在某个点上程序给出了可预测的结果,然后函数返回一个非常接近 unsigned的数字long long int 范围。它发生的点取决于读取的值,例如,如果数字串主要由8s和9s组成,它将比它只有5s和6s时更早发生。为什么会这样?

#include <iostream>
#include <fstream>

using namespace std;


int product (int res, int a, char buffer[]){
for (int i = 0; i < a; i++){
//simple char to int conversion
res*=(buffer[i] - '0');
}

return res;
}

int main () {
char check;
int res = 1;
fstream plik;
plik.open ("8.txt");
unsigned long long int high;
unsigned long long int result;
//main function in the program
if (plik.good()){
    char buffer [13];
    for (int i = 0; i < 13; i++){
        plik >> buffer[i];
    }
    result = product (res, 13, buffer);
    high = result;
    cout << high << endl;
    //the main checking loop
    while (!plik.eof()){
    //just an interruption to make it possible to view consecutive products
    //the iteration in the buffer
    for (int i = 0; i < 12; i++){
    buffer[i] = buffer[i+1];
    }
    plik >> buffer[12];
    result = product (res, 13, buffer);
    //comparison between the current product and highest one
    if (high < result){
    high = result;
    }
    cin >> check;
    cout << high << endl;
    //again a tool for checking where the problem arises
    for (int i = 0; i < 13; i++){
        cout << buffer[i] << "  ";
    }
    cout << endl;
    }
    plik.close();
    cout << high << endl;
}

return 0;

}

该程序打印出当前最高的产品和当前包含在阵列中的所有数字。 它看起来像这样: The error

1 个答案:

答案 0 :(得分:1)

使用unsigned long long int代替int来计算产品。 13位数的乘积很容易变得大于最大的int。