uva 10931 - Parity

最新推荐文章于 2019-04-29 13:34:38 发布

sunnykema

最新推荐文章于 2019-04-29 13:34:38 发布

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分类专栏：算法设计实验编程 uva

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本文介绍了一个简单的算法，用于计算任意给定整数（范围1至2^31-1）的二进制表示中1的个数的奇偶性，并通过示例输入输出展示了如何使用该算法。

Parity

Time Limit: 1 second

We define the parity of an integer n as the sum of the bits in binary representationcomputed modulo two. As an example, the number 21 = 10101₂ has three 1sin its binary representation so it has parity 3 (mod 2), or 1.

In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647.

Input

Each line of the input has an integer I and the end of the input is indicated by a linewhere I = 0 that should not be processed.

Output

For each integer I in the inputt you should print a line The parity of B is P (mod 2).,where B is the binary representation of I.

Sample Input

Sample Output

The parity of 1 is 1 (mod 2).
The parity of 10 is 1 (mod 2).
The parity of 1010 is 2 (mod 2).
The parity of 10101 is 3 (mod 2).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>

using namespace std;
#define maxn 10005
#define inf 0x7ffffff
pair <int ,int > p;
int arr[maxn];
int n;
//int cmpD(Node a,Node b){
//    return a.h > b.h;
//}
//int cmpI(Node a,Node b){
//    return a.h < b.h;
//}
int main()
{
    while(scanf("%d",&n) != EOF && n){
        int tmp = n;
        int cnt = 0;
        int ans = 0;
        while(tmp){
            arr[cnt] = tmp % 2;
            if(arr[cnt] == 1){
                ans ++;
            }
            tmp /= 2;
            cnt++;
        }
        printf("The parity of ");
        for(int i = cnt - 1; i >= 0; i--){
            printf("%d",arr[i]);
        }
        printf(" is %d (mod 2).\n",ans);
    }
    return 0;
}