英文:
Multiplication of numbers from a to b
问题
我需要编写一个算法，用于在不需要输入（scanf）的情况下计算从a到b的数的乘积。就像这样：
a = 2;
b = 6;
2 * 3
2 * 4
...
2 * 6
我有自己的算法：
void main()
{
int dist = 1;
int a = 2;
int b = 5;
for (int i = a; a <= b; a++) {
printf("%d", a * a++);
}
}
但它不正确。
英文:
I need to write algorithm, that multiplies numbers from a to b without input (scanf). Like this:
a = 2;
b = 6;
2 * 3
2 * 4
...
2 * 6
I have my algorithm:
void main()
{
int dist = 1;
int a = 2;
int b = 5;
for (int i = a; a <= b; a++) {
printf("%d", a * a++);
}
}
but it doesn't work correct
答案1
得分: 5
这是因为你在上面的示例中对 a 进行了两次增加 (a++)。同时你有点混淆了 a 和 i。正确的代码如下：
int a = 2;
int b = 5;
for (int i = a; i <= b; i++)
{
printf("%d * %d = %d\n", a, i, a * i);
}
它会输出：
> 2 * 2 = 4
>
> 2 * 3 = 6
>
> 2 * 4 = 8
>
> 2 * 5 = 10
英文:
This is because you are increasing a (a++) two times in your example above. Also you mixed up a and i a little bit. Correct one is:
int a = 2;
int b = 5;
for (int i = a; i <= b; i++)
{
printf("%d * %d = %d\n", a, i, a * i);
}
which prints:
> 2 * 2 = 4
>
> 2 * 3 = 6
>
> 2 * 4 = 8
>
> 2 * 5 = 10
答案2
得分: 0
int main() {
int a = 2;
int b = 6;
int result = 1; // 初始化结果变量为1以进行乘法运算
// 从a到b执行乘法运算
for (int i = a; i <= b; i++) {
result *= i;
}
printf("从%d到%d的乘法结果是：%d\n", a, b, result);
return 0;
}
英文:
int main() {
int a = 2;
int b = 6;
int result = 1; // Initialize the result variable to 1 for multiplication
// Perform multiplication from a to b
for (int i = a; i <= b; i++) {
result *= i;
}
printf("The multiplication result from %d to %d is: %d\n", a, b, result);
return 0;
}
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