Assume a=10, b=23.5, please solve a+b, a-b, a*b, a/b? Assume: a=3,b=4,c=5, please solve a==b!=c and a!=b==c 用C语言写出来
时间: 2024-06-14 20:06:56 浏览: 11
以下是使用C语言解决这些问题的代码示例:
1. 解决a+b, a-b, a*b, a/b:
```c
#include <stdio.h>
int main() {
int a = 10;
float b = 23.5;
printf("a + b = %f\n", a + b);
printf("a - b = %f\n", a - b);
printf("a * b = %f\n", a * b);
printf("a / b = %f\n", a / b);
return 0;
}
```
2. 解决a==b!=c和a!=b==c:
```c
#include <stdio.h>
int main() {
int a = 3;
int b = 4;
int c = 5;
if (a == b && b != c) {
printf("a == b != c\n");
} else if (a != b && b == c) {
printf("a != b == c\n");
} else {
printf("Neither a == b != c nor a != b == c\n");
}
return 0;
}
```
相关问题
a=2;b=3;a=b+4;a=
This code is not valid.
If we assume that there is a semicolon after "b=3", then the code would be:
a=2;b=3;a=b;4;a=
In this case, the number 4 is not assigned to any variable and is not a valid statement. It is possible that the intention was to assign 4 to a variable, such as:
a=2;b=3;a=b;num=4;a=
However, the last line is incomplete and cannot be executed.
How many bits would I need to write down an integer a in base b?
If we assume that the integer a has n digits when written in base b, then we can say that a is in the range b^(n-1) <= a < b^n. Taking the logarithm base 2 of both sides, we get:
log2(b^(n-1)) <= log2(a) < log2(b^n)
(n-1)log2(b) <= log2(a) < nlog2(b)
Since n-1 < n, we can use n in place of n-1 in the first inequality to get:
nlog2(b) <= log2(a) < nlog2(b)
Subtracting the left-hand side from the right-hand side, we get:
nlog2(b) - nlog2(b) <= log2(a) - nlog2(b) < (n+1)log2(b) - nlog2(b)
Simplifying, we get:
0 <= log2(a) - nlog2(b) < log2(b)
This tells us that the difference between log2(a) and nlog2(b) is less than log2(b), or equivalently, that nlog2(b) is within log2(b) of log2(a). Therefore, we can say that:
n = ceil(log2(a)/log2(b))
This formula gives us the minimum number of digits we need to represent a in base b. Since each digit can take on b possible values (0 through b-1), we need log2(b) bits to represent each digit. Therefore, the total number of bits required to represent a in base b is:
ceil(log2(a)/log2(b)) * log2(b)
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