Input an integer[-2147483648,+2147483647],output the binary of it
时间: 2023-04-04 22:01:24 浏览: 91
好的,我可以回答这个问题。将输入的整数转换为二进制,可以使用位运算符和循环来实现。以下是一个示例代码:
```
#include <stdio.h>
void printBinary(int num) {
for (int i = 31; i >= ; i--) {
if (num & (1 << i)) {
printf("1");
} else {
printf("");
}
}
}
int main() {
int num;
printf("请输入一个整数:");
scanf("%d", &num);
printf("%d的二进制表示为:", num);
printBinary(num);
printf("\n");
return ;
}
```
当输入一个整数时,程序会输出该整数的二进制表示。例如,输入10,输出101。
相关问题
Write a javascript function that takes an integer as input, and returns the number of bits that are equal to one in the binary representation of that number. You can guarantee that input is non-negative. Example: The binary representation of 1234 is 10011010010, so the function should return 5 in this case
Sure, here's a JavaScript function that takes an integer as input and returns the number of bits that are equal to one in its binary representation:
```javascript
function countBits(num) {
// Convert the number to binary string
let binary = num.toString(2);
// Count the number of ones in the binary string
let count = 0;
for (let i = 0; i < binary.length; i++) {
if (binary[i] === '1') {
count++;
}
}
return count;
}
```
You can call this function with an integer as input, like this:
```javascript
console.log(countBits(1234)); // Output: 5
```
This will output `5`, which is the number of bits that are equal to one in the binary representation of `1234`.
Once upon a time, Toma found himself in a binary cafe. It is a very popular and unusual place. The cafe offers visitors k different delicious desserts. The desserts are numbered from 0 to k−1 . The cost of the i -th dessert is 2i coins, because it is a binary cafe! Toma is willing to spend no more than n coins on tasting desserts. At the same time, he is not interested in buying any dessert more than once, because one is enough to evaluate the taste. In how many different ways can he buy several desserts (possibly zero) for tasting? Input The first line of the input contains a single integer t (1≤t≤1000 ) — the number of test cases. Then follows t lines, each of which describes one test case. Each test case is given on a single line and consists of two integers n and k (1≤n,k≤109 ) — the number of coins Toma is willing to spend and the number of desserts in the binary cafe. Output Output t integers, the i -th of which should be equal to the answer for the i -th test case — the number of ways to buy desserts for tasting.
This is a classic problem of finding the number of subsets of a given set with a certain property. In this case, we want to find the number of subsets of the set {0,1,2,...,k-1} such that the sum of the elements in the subset is less than or equal to n.
We can use dynamic programming to solve this problem efficiently. Let dp[i][j] be the number of ways to choose a subset of the first i elements of the set {0,1,2,...,k-1} such that the sum of the elements in the subset is exactly j. Then the answer to the problem is the sum of dp[k][0] to dp[k][n].
The base case is dp[0][0]=1, since there is exactly one way to choose an empty subset with sum zero. For each i from 1 to k, we can either choose the i-th element or not. If we choose it, then we need to find the number of ways to choose a subset of the first i-1 elements with sum j-2i. If we do not choose it, then we need to find the number of ways to choose a subset of the first i-1 elements with sum j. Therefore,
dp[i][j] = dp[i-1][j] + dp[i-1][j-2i], if j>=2i
dp[i][j] = dp[i-1][j], otherwise
The final answer is the sum of dp[k][0] to dp[k][n]. The time complexity of this algorithm is O(kn).
Here is the Python code to solve the problem:
```python
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
dp = [[0] * (n+1) for _ in range(k+1)]
dp[0][0] = 1
for i in range(1, k+1):
for j in range(n+1):
dp[i][j] = dp[i-1][j]
if j >= 2**i:
dp[i][j] += dp[i-1][j-2**i]
print(sum(dp[k]))
```
I hope this helps! Let me know if you have any more questions.
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