C++ Project Euler 8

C++ Project Euler 8

本文关键字:Euler Project C++      更新时间:2023-10-16

问题:找到连续数字的最大产品

如果我们试图找到连续4个数字的最大产品,但我的代码有效,但是当我尝试找到13时给我一个错误的答案。

4个相邻数字的最大乘积为= 5832

#include<iostream>
#include<string>
using namespace std;
int main() {

    //Convert String array to char array
    char x [] = { "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450" };
    int y[1000];
    for (unsigned int i = 0; i <=999; i++) {
        y[i] = x[i] - 48;
    }
    long long max = y[0] * y[1] * y[2] * y[3];//* y[4] * y[5] * y[6] * y[7] * y[8] * y[9] * y[10] * y[11] * y[12];
    cout << max << endl;
    long long  current = 0;
    for (unsigned int i = 1; i <= 999; i++) {
        current = y[i] * y[i + 1] * y[i + 2] * y[i + 3];// * y[i+4] * y[i+5] * y[i+6] * y[i+7] * y[i+8] * y[i+9] * y[i+10] * y[i+11] * y[i+12];     
        if (current > max) {
            max = current;
        }
    }
    cout <<"Max is  " <<  max << endl;
}
for (unsigned int i = 1; i <= 999; i++) {
        current = y[i] * y[i + 1] * y[i + 2] * y[i + 3];// * y[i+4] * y[i+5] * y[i+6] * y[i+7] * y[i+8] * y[i+9] * y[i+10] * y[i+11] * y[i+12];     
        if (current > max) {
            max = current;
        }
    }

说何时i = 999,您认为y[i+1]会给您什么?提示:y足够大吗?y可以在索引1000中具有元素吗?

简单的代码修复且有效。

long long y [1000];