欧拉项目第8题：13位相邻数字最大乘积C++代码问题排查

问题核心原因 #

你的代码存在三个关键问题，导致结果与正确值不符：

1. 整数溢出
   int类型的最大值通常为2147483647，而13个9的乘积为9^13 = 2541865828329，远超出int的存储范围。用int存储乘积会触发溢出，数值被截断，最终得到错误结果。

2. 未完整实现13个数字的乘积计算
   当前代码仅计算了4个相邻数字的乘积，后续9个数字的乘积部分被注释，无法完成13个数字的乘积计算。

3. 循环边界错误
   循环条件i < size(v1) -13会漏掉最后一组13个数字的组合。正确的边界应为i <= size(v1)-13（或i < size(v1)-12），确保所有合法的起始位置都被遍历。

修正后的代码 #

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> v1 = {7,3,1,6,7,1,7,6,5,3,1,3,3,0,6,2,4,9,1,9,2,2,5,1,1,9,6,7,4,4,2,6,5,7,4,7,4,2,3,5,5,3,4,9,1,9,4,9,3,4,
                      9,6,9,8,3,5,2,0,3,1,2,7,7,4,5,0,6,3,2,6,2,3,9,5,7,8,3,1,8,0,1,6,9,8,4,8,0,1,8,6,9,4,7,8,8,5,1,8,4,3,
                      8,5,8,6,1,5,6,0,7,8,9,1,1,2,9,4,9,4,9,5,4,5,9,5,0,1,7,3,7,9,5,8,3,3,1,9,5,2,8,5,3,2,0,8,8,0,5,5,1,1,
                      1,2,5,4,0,6,9,8,7,4,7,1,5,8,5,2,3,8,6,3,0,5,0,7,1,5,6,9,3,2,9,0,9,6,3,2,9,5,2,2,7,4,4,3,0,4,3,5,5,7,
                      6,6,8,9,6,6,4,8,9,5,0,4,4,5,2,4,4,5,2,3,1,6,1,7,3,1,8,5,6,4,0,3,0,9,8,7,1,1,1,2,1,7,2,2,3,8,3,1,1,3,
                      6,2,2,2,9,8,9,3,4,2,3,3,8,0,3,0,8,1,3,5,3,3,6,2,7,6,6,1,4,2,8,2,8,0,6,4,4,4,4,8,6,6,4,5,2,3,8,7,4,9,
                      3,0,3,5,8,9,0,7,2,9,6,2,9,0,4,9,1,5,6,0,4,4,0,7,7,2,3,9,0,7,1,3,8,1,0,5,1,5,8,5,9,3,0,7,9,6,0,8,6,6,
                      7,0,1,7,2,4,2,7,1,2,1,8,8,3,9,9,8,7,9,7,9,0,8,7,9,2,2,7,4,9,2,1,9,0,1,6,9,9,7,2,0,8,8,8,0,9,3,7,7,6,
                      6,5,7,2,7,3,3,3,0,0,1,0,5,3,3,6,7,8,8,1,2,2,0,2,3,5,4,2,1,8,0,9,7,5,1,2,5,4,5,4,0,5,9,4,7,5,2,2,4,3,
                      5,2,5,8,4,9,0,7,7,1,1,6,7,0,5,5,6,0,1,3,6,0,4,8,3,9,5,8,6,4,4,6,7,0,6,3,2,4,4,1,5,7,2,2,1,5,5,3,9,7,
                      5,3,6,9,7,8,1,7,9,7,7,8,4,6,1,7,4,0,6,4,9,5,5,1,4,9,2,9,0,8,6,2,5,6,9,3,2,1,9,7,8,4,6,8,6,2,2,4,8,2,
                      8,3,9,7,2,2,4,1,3,7,5,6,5,7,0,5,6,0,5,7,4,9,0,2,6,1,4,0,7,9,7,2,9,6,8,6,5,2,4,1,4,5,3,5,1,0,0,4,7,4,
                      8,2,1,6,6,3,7,0,4,8,4,4,0,3,1,9,9,8,9,0,0,0,8,8,9,5,2,4,3,4,5,0,6,5,8,5,4,1,2,2,7,5,8,8,6,6,6,8,8,1,
                      1,6,4,2,7,1,7,1,4,7,9,9,2,4,4,4,2,9,2,8,2,3,0,8,6,3,4,6,5,6,7,4,8,1,3,9,1,9,1,2,3,1,6,2,8,2,4,5,8,6,
                      1,7,8,6,6,4,5,8,3,5,9,1,2,4,5,6,6,5,2,9,4,7,6,5,4,5,6,8,2,8,4,8,9,1,2,8,8,3,1,4,2,6,0,7,6,9,0,0,4,2,
                      2,4,2,1,9,0,2,2,6,7,1,0,5,5,6,2,6,3,2,1,1,1,1,1,0,9,3,7,0,5,4,4,2,1,7,5,0,6,9,4,1,6,5,8,9,6,0,4,0,8,
                      0,7,1,9,8,4,0,3,8,5,0,9,6,2,4,5,5,4,4,4,3,6,2,9,8,1,2,3,0,9,8,7,8,7,9,9,2,7,2,4,4,2,8,4,9,0,9,1,8,8,
                      8,4,5,8,0,1,5,6,1,6,6,0,9,7,9,1,9,1,3,3,8,7,5,4,9,9,2,0,0,5,2,4,0,6,3,6,8,9,9,1,2,5,6,0,7,1,7,6,0,6,
                      0,5,8,8,6,1,1,6,4,6,7,1,0,9,4,0,5,0,7,7,5,4,1,0,0,2,2,5,6,9,8,3,1,5,5,2,0,0,0,5,5,9,3,5,7,2,9,7,2,5,
                      7,1,6,3,6,2,6,9,5,6,1,8,8,2,6,7,0,4,2,8,2,5,2,4,8,3,6,0,0,8,2,3,2,5,7,5,3,0,4,2,0,7,5,2,9,6,3,4,5,0,};

    // 使用long long存储乘积，避免溢出
    long long max_product = 0;
    int i = 0;
    // 修正循环边界，遍历所有合法的起始位置
    while (i <= v1.size() - 13) {
        long long multiplication = 1;
        // 计算13个相邻数字的乘积
        for (int j = 0; j < 13; j++) {
            multiplication *= v1[i + j];
        }
        if (multiplication > max_product) {
            max_product = multiplication;
        }
        i++;
    }
    cout << "The greatest product is: " << max_product << endl;

    return 0;
}

额外优化建议 #

• 可以跳过包含0的组合，因为只要有一个数字是0，乘积就是0，无需计算，能提升效率。
• 避免使用using namespace std;，在大型项目中容易引发命名冲突。

内容的提问来源于stack exchange，提问作者Chapster