爲什麼我的程序在某些測試中正常工作，而不是爲其他人測試？

Question

這是我的測試輸出：

:) greedy exists 

:) greedy compiles 

:(input of 0.41 yields output of 4 
    expected "4\n", not "3\n" 

:(input of 0.01 yields output of 1 
    expected "1\n", not "0\n" 

:) input of 0.15 yields output of 2 

:) input of 1.6 yields output of 7 

:) input of 23 yields output of 92 

:) input of 4.2 yields output of 18 

:) rejects a negative input like -.1 

:) rejects a non-numeric input of "foo" 

:) rejects a non-numeric input of "" 

這是代碼：

#include <stdio.h> 
#include <cs50.h> 

void count_coins(); 

int coin = 0; 
float change = 0; 

int main(void) 
{ 
    do 
    { 
    change = get_float("How much change is owed? "); 
    } 
    while (change < 0); 

    count_coins(); 

    printf("%i\n", coin); 
} 

void count_coins() 
{ 
    while (change > 0.24) 
    { 
     coin++; 
     change -= 0.25; 
     // printf("%.2f\n", change); 
    } 

    while (change > 0.09) 
    { 
     coin++; 
     change -= 0.10; 
     // printf("%.2f\n", change); 
    } 

    while(change > 0.04) 
    { 
     coin++; 
     change -= 0.05; 
     // printf("%.2f\n", change); 
    } 

    while (change >= 0.01) 
    { 
     coin++; 
     change -= 0.01; 
     // printf("%.2f\n", change); 
    } 
} 

Answer 1

由於Havenard已經寫了，問題是，change被存儲爲float。 對於這樣的程序，change必須存儲爲整數值。

這是你與int，而不是float代碼：

#include <stdio.h> 
#include <cs50.h> 

void count_coins (void); 

int coin = 0; 
int change = 0; 

int main (void) { 
    do { 
     change = 41; // insert a get integer function here 
    } while (change < 0); 

    count_coins(); 

    printf("%i\n", coin); 
} 

void count_coins (void) { 
    while (change >= 25) { 
     coin++; 
     change -= 25; 
    } 

    while (change >= 10) { 
     coin++; 
     change -= 10; 
    } 

    while(change >= 5) { 
     coin++; 
     change -= 5; 
    } 

    while (change >= 1) { 
     coin++; 
     change -= 1; 
    } 
}