如何将num String中的每个数字转换为C中的int

Question

我一直在尝试在 Project Euler

上解决这个问题

这是我到目前为止所做的 -

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
    char nums[] = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
    int64_t sum = 1, new_sum = 0;
    int num;
    for(int i = 0; i < 988; i+= 13){
        sum = 1
        for(int j = 0; j < 13; ++j){
            //Converting nums[i] to int and storing it to num here
            sum *= num;
        }
        if(sum > new_sum){
            new_sum = sum;
        }
    }
    printf("%lld", new_sum);
}

我不知道如何将字符串中的每个 charNum 转换为整数，我试过 atoi 和 sscanf，它们都要求一个字符指针。

我分别得到这些错误

passing argument to parameter here
int      atoi(const char *); //sum *= atoi(nums[i])

format specifies type 'int *' but the argument has type 'int' [-Wformat]
            sscanf(nums[i], "%d", num);

感谢任何帮助。

Answer 1

我假设您希望每个数字都有一个数字（即范围 0 到 9）如果是，下面的代码应该与您的需要相差不远：

char nums[] = "73167176531330624919225119674426574742355349194934969835203... (so long)...";

char *p; 
long lTheTotalCount, lTheNumber;

lTheTotalCount = 0;
for (p=nums ; *p ; p++) {
   lTheNumber = (long)(*p - '0');
   lTheTotalCount += lTheNumber;
};

Answer 2

请您尝试以下操作：

#include <stdio.h>
#include <stdlib.h>
#define N 13

int main() {
    char nums[] = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
    int64_t prod = 1, max_prod = 0;
    int num;
    int pos;
    for (int i = 0; i < sizeof nums; i += N) {
        prod = 1;
        for (int j = 0; j < N; j++){
            // Converting nums[i] to int and storing it to num here
            num = nums[i + j] - '0';
            prod *= num;
        }
        if (prod > max_prod) {
            max_prod = prod;
            pos = i;
        }
    }
    printf("max product = %ld at %.*s\n", max_prod, N, nums + pos);
}

输出：

max product = 6270566400 at 4355766896648

顺便说一句，因为变量名称 sum 具有误导性，我已将其更改为产品的 prod。

Answer 3

只需将表示数字的字符转换为数字本身即可，

int digit = nums[position] - '0';

C 语言保证 any 字符集的数字 0-9 以完全相同的顺序依次出现（字母字符不一定是这种情况，参见例如（在-？）著名的 EBCDIC）。

当我读到这个问题时，最大值可以在任何位置，而不仅仅是 13 的倍数。这也说明了四个后续数字的最大值位于前面的位数是不是四的倍数的位置。

除此之外，除非数字是 13 的倍数，否则您的循环将超出 nums 的数组边界，因此未定义的行为（如果不是 13 的倍数，则 [= 的最后一次迭代13=] 剩下不到 13 位数字，但 j 仍然尝试遍历所有 13 位数字）。

修复两者：

size_t sequenceLength = 13; // could be a function parameter

size_t len = strlen(nums); // sizeof would include the terminating 0 character!

if(len < sequenceLength)
{
    return 0; // if within a function, else whatever appropriate error handling
}

uint64_t max = 1; // can't have negative values anyway...
for(size_t i = 0; i < sequenceLength; ++i)
{
    max *= nums[i] - '0'; // see above
}
// now we found first possible maximum: the product of first 13 digits

uint64_t num = max;
for(size_t i = sequenceLength; i < len; ++i)
{
    // for next product, we have already included the first 12 digits
    // within the previous product!

    // but there is one surplus digit contained we need to eliminate:
    num /= nums[i - sequenceLength] - '0';

    // now we can include the yet missing one:
    num *= nums[i] - '0';

    // or as a one-liner:
    num = num / (nums[i - sequenceLength] - '0') * (nums[i] - '0');

    // TODO: update maximum, if need be, just as in your code
}

如何将num String中的每个数字转换为C中的int

How to convert each number in a num String to an int in C

c

string

type-conversion