11209: 【原1209】数数

题目描述

author: Qinglin Li 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1209

Description

Little Jealous Joker是个比较“2”的人,特别喜欢2进制。有一天，Little Jealous从书中翻出来N个数a(1),a(2),...,a(n)，他希望把这些数全部转化为2进制，数一数里面一共有多少个1。由于Little Jealous Joker已经“2”到不会数数了，请你帮帮他吧。

Input Format

第一行一个数N 第二行N个数a(1), a(2), ..., a(n)，以空格隔开

Output Format

一个数，1的总个数

Sample Input

3
3 4 5

Sample Output

Limits

n<=100,000 a(i)不超过32位带符号整数(int)的范围

FineArtz's solution

/* 数数 */
#include <iostream>
using namespace std;

int main(){
   int n, ans = 0;
   cin >> n;
   for (int i = 1; i <= n; ++i){
      int a = 0;
      cin >> a;
      while (a > 0){
         if (a % 2) ++ans;
         a /= 2;
      }
   }
   cout << ans << endl;
   return 0;
}

ligongzzz's solution

#include "iostream"
#include "cstdio"
using namespace std;

int main() {
    int num;
    cin >> num;

    long long ans = 0;

    for (int i = 0; i < num; ++i) {
        int temp;
        scanf("%d", &temp);
        //转二进制
        for (; temp > 0; temp = temp >> 1)
            if (temp % 2 == 1)
                ++ans;
    }

    cout << ans;

    return 0;
}

q4x3's solution

#include <iostream>
#define POW(c) (1 << (c))
#define MASK(c) (((unsigned long) -1) / (POW(POW(c)) + 1))
#define ROUND(n, c) (((n) & MASK(c)) + ((n) >> POW(c) & MASK(c)))

using namespace std;

int main()
{
    int n, ans = 0;
    cin >> n;
    for (int i = 1;i <= n;++ i) {
        int num;
        cin >> num;
        num = ROUND (num, 0);
        num = ROUND(num, 1);
        num = ROUND(num, 2);
        num = ROUND(num, 3);
        num = ROUND(num, 4);
        ans += num; // O(loglognum)
        /*while (0 < num) {
            ++ ans;
            num &= num - 1;
        }*/ // O(ones)
        /*while (0 < num) {
            ans += (1 & num);
            num >>= 1;
        }*/ // O(lognum)
    }
    cout << ans << endl;
}

skyzh's solution

#include <iostream>
using namespace std;

int main() {
    int n;
    long long total = 0;
    cin >> n;
    for (int i = 0; i < n; i++) {
        unsigned int f;
        cin >> f;
        while (f > 0) {
            total += f & 1;
            f >>= 1;
        }
    }
    cout << total << endl;
    return 0;
}

victrid's solution

#include <iostream>
using namespace std;
int main()
{
    int N;
    cin >> N;
    int num;
    int cnt = 0;
    for (int i = 0; i < N; i++)
    {
        cin >> num;
        while (num != 0)
        {
            if (num & 1)
                cnt++;
            num = num >> 1;
        }
    }
    cout << cnt;
    return 0;
}

yyong119's solution

#include <iostream>
using namespace std;
int main() {

    ios::sync_with_stdio(false);
    int n, total = 0;
    cin >> n;
    while (n--) {
        long tmp;
        cin >> tmp;
        if (tmp < 0) tmp = -tmp;
        while (tmp) {
            total += tmp % 2;
            tmp /= 2;
        }
    }
    cout << total << endl;
    return 0;
}