c - 在 C 中实现求和公式

Question

所以，我需要在 C 中实现这个公式:

其中 n 由用户输入，必须 >= 1。Ci 是 n 和 k 的第 i 个数字> 数字，Ci 与线是 Ci 的补码(数字 + 补码 = 9)。例如，数字 21262 满足此等式，因为:

21262 = 7^5 + 8^4 + 7^3 + 3^2 + 7^1 = 16807 + 4096 + 343 + 9 + 7 = 21262

我试过制作一个算法并将其转换为 C 程序，但是当我执行时，出现了问题。我在制定执行 k-i+1 幂的循环时遇到问题。

#include <stdio.h>

int main() {
int n, t, k, c, i, s, power, j;
do {
    printf("Enter n = ");
    scanf("%d", &n);
} while (n < 1);

t = n;                    // Here we save the value of n, and operate with t
k = 0;                          // k counts the number of digits 
while (t > 0) {
    t = t / 10;      
    k++;                        // k = number of digits
}                                
t = n;
s = 0;

for (i = 1; i <= k; i++) {            // Starts the sum from i to k
    c = 9 - t % 10;                   // Complements the digits
    power = 1;

    for (j = 1; j <= k-i+1; j++)      // Start of loop that powers th number
        power = c * power;
    s = s + power;
    t = t / 10;
}

if (s == n)
    printf("The number fulfills the equation");
else
    printf("The number doesn't fulfill the equation");
return 0;
}

如您所见，我尝试通过制作一个将 c 补码自身乘以 k-i+1 次的循环来解决幂问题。但是有些不对劲。请帮忙!

Answer 1

当您从右到左阅读时，幂应该是 i 而不是 k - i + 1

#include<stdio.h>


int main() {
    int n, t, k, c, i, s, power, j;
    do {
        printf("Enter n = ");
        scanf("%d", &n);
    } while (n < 1);

    t = n;                    // Here we save the value of n, and operate with t
    k = 0;                          // k counts the number of digits
    while (t > 0) {
        t = t / 10;
        k++;                        // k = number of digits
    }
    t = n;
    s = 0;

    for (i = 1; i <= k; i++) {            // Starts the sum from i to k
        c = 9 - (t % 10);                   // Complements the digits
        power = 1;

        for (j = 1; j <= i; j++)      // Start of loop that powers th number
            power = c * power;
        s = s + power;
        t = t / 10;
    }

    if (s == n)
        printf("The number fulfills the equation");
    else
        printf("The number doesn't fulfill the equation");
    return 0;
}