我在下面写的程序似乎正在完成它的工作，但它产生了错误的答案。如评论中所述，我需要将1000位数字符串的5个连续数字相乘，然后在整个字符串中找到最大的产品。
/*
OBJECTIVE: Multiply five consecutive numbers and then compare products to find largest product
PROVE: 40824
*/
#include <iostream>
using namespace std;
int main()
{
// string length is 1000 characters
string str = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
int store = 0; // what I need to compare total to later
int total = 1; // total must be '1' for multiplication or result is always '0'
for(int i = 0;i<=999;) // i = 0 because that's the starting point for place of first number in string (str[0]); because the length is 1 short, 999 is the limit
{
total = 1; // what will be multiplied by inputs from str[]
int incr = i + 4; // places for the next 5 now that 0-4 are out of the way would be 5,6,7,8,9
// A general note: pattern this generates is that the end digit of an even break in groupings will always either be 4 or 9
// e.g. breaking the loop at position 44, 199, 234 will multiply 5 places properly without error
if(i == 0) // when starting out i = 0
{
for(;i<=4;i++) // i = str[0] which is then incrmemented to str[4] which is fifth number
{
cout << "i: " << i << '\t' << "str[" << i << "]: " << str[i] << '\t';
total*=(str[i] - '0'); // takes str[1],str[2],...str[4] and multiplies them together using total as valueholder.
// " - '0' " is because string numbers return ASCII codes
cout << "Total Product: " << total << '\t';
cout << "Store: " << store << '\t' << endl;
};
store = total > store ? total:store; // if total is greater than store, then store = total; simple enough
cout << endl; // line-break after first set of five
};
if(i>=5) // i is no longer 0
{
for(;i<=incr;i++) // i must be added by 5 now because 0,1,2,3,4 are first five and 5,6,7,8,9 are next
{
cout << "i: " << i << '\t' << "str[" << i << "]: " << str[i] << '\t';
total*=(str[i] - '0'); // takes str[1],str[2],...str[4] and multiplies them together using total as valueholder.
// " - '0' " is because string numbers return ASCII codes
cout << "Total Product: " << total << '\t';
cout << "Store: " << store << '\t' << endl;
};
store = total > store ? total:store; // if total is greater than store, then store = total; simple enough
cout << endl; // line-break after each set of five
};
};
cout << endl << endl << store; // final output; what should be answer
return 0;
}
该程序显示它确实将五个连续数字相乘并在必要时成功提升了商店价值，唉我仍然得到31752的错误答案，应该是40824（根据项目Euler Solutions google-group：{{ 3}}）。我做错了什么？我该怎么做才能修复这个程序？未来的任何提示，以避免我没有看到的可能是愚蠢的错误？
答案 0 :(得分：1)
从粗略的一瞥，我相信你的问题在于你在i循环声明之外的多个位置修改for。这将导致您扫描数字以跳过可能正确的序列。\
如果你推理出你的代码，它会从字符串中的第0位开始。然后你有一个条件，你增加i 4次，所以然后下一次循环，你开始尝试从字符串中的位置4开始寻找一个序列，而不是从位置1开始。
答案 1 :(得分：1)
固定代码：
/*
OBJECTIVE: Multiply five consecutive numbers and then compare products to find largest product
PROVE: 40824
*/
#include <iostream>
using namespace std;
int main()
{
// string length is 1000 characters
string str = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
int store = 0; // what I need to compare total to later
int total = 1; // what will be multiplied by inputs from str[]
int n = 0; // keeps track of single iterations
for(int i = 0;n<=999;n++) // i = 0 because that's the starting point for place of first number in string (str[0]); because the length is 1 shorter, 999 is the limit
{
total = 1; // must be set back to one each successful product loop
int incr = n + 4; // places for the next 5 after 0
// A general note: pattern this generates is that the end digit of an even break in groupings will always either be 4 or 9
// e.g. breaking the loop at position 44, 199, 234 will multiply 5 places properly without error
if(n == 0) // string starts at 0, must be accounted for especially
{
for(;i<=4;i++) // i = str[0] which is then incrmemented to str[4] which is fifth number
{
cout << "n: " << n << '\t' << "str[" << i << "]: " << str[i] << '\t';
total*=(str[i] - '0'); // takes str[1],str[2],...str[4] and multiplies them together using total as valueholder.
// " - '0' " is because string numbers return ASCII codes
cout << "Total Product: " << total << '\t';
cout << "Store: " << store << '\t' << endl;
};
store = total > store ? total:store; // if total is greater than store, then store = total; simple enough
};
for(;i<=incr;i++) // i must be added by 4 now because 0,1,2,3,4 are first five and 1,2,3,4,5 are next
{
cout << "n: " << n << '\t' << "str[" << i << "]: " << str[i] << '\t';
total*=(str[i] - '0'); // takes str[1],str[2],...str[4] and multiplies them together using total as valueholder.
// " - '0' " is because string numbers return ASCII codes
cout << "Total Product: " << total << '\t';
cout << "Store: " << store << '\t' << endl;
};
store = total > store ? total:store; // if total is greater than store, then store = total; simple enough
i = (n + 1); // i needs to start at a consecutive n each successful loop
};
cout << endl << endl << store;
return 0;
}
答案 2 :(得分：1)