Question

任何人都可以帮我处理我的代码

int main(void){

int money, x[6],i[6];

x[0] = 1000; x[1] = 500; x[2] = 200; x[3] = 100; x[4] = 50; x[5] = 20;

printf("Enter your Money: ");
scanf("%d", &money);

printf("\nBreakdown:\n");

for(int y=0;money != 0; y++){

    i[y] = ( money - (money % x[y]) )/x[y];
    money = (money % x[y]);

    printf("%10d  x  %d  =  %5d \n", x[y], i[y], (x[y] * i[y]));

}
printf("---------------------------\n");
printf("Total:");
printf("                  %d", money);

的getch（）;

}

如何检查是否有余数＆amp;它会说输入无效

Answer 1

这是一个适用于所有输入的版本，而不仅仅是可被20整除的版本。

int main(void){

    int money, money2, x[7],i[7], y;

    x[0] = 1000; x[1] = 500; x[2] = 200; x[3] = 100; x[4] = 50; x[5] = 20, x[6] = 1;

    printf("Enter your Money: ");
    scanf("%d", &money);
    money2 = money;

    printf("\nBreakdown:\n");

    for(y=0;money >0; y++){

        i[y] = ( money - (money % x[y]) )/x[y];
        money = (money % x[y]);

        printf("%10d  x  %d  =  %5d \n", x[y], i[y], (x[y] * i[y]));

    }
    printf("---------------------------\n");
    printf("Total:");
    printf("                  %d", money2);

    getch();
}

所做的更改

数组大小增加到7.并添加x[6] = 1。这确保了最终案例的处理。
添加了一个额外的变量money2来存储最后用于打印的总金额。在您的代码中，money变量在到达打印函数时会更改并且值为零。

Answer 2

你可以为x [6] = 5，x [7] = 2，x [8] = 1等数组添加更多值，我认为这是最好的想法。
这样你就可以计算出5,2和1的倍数否则

你可以做这样的事情。

int nTemp = money; //Take a temp variable to store total money 
for(int y=0;money != 0 && y<6; y++){ // check both money!=0 and y should be less than 6

    i[y] = ( money - (money % x[y]) )/x[y];
    money = (money % x[y]);

    printf("%10d  x  %d  =  %5d \n", x[y], i[y], (x[y] * i[y]));
}
if(money>0) // if remainder is there you can print
   printf("Remaining change = %d\n", money);

printf("---------------------------\n");
printf("Total:");
printf("                  %d", nTemp); // print nTemp here, i.e total money

C中的货币面额

2 个答案: