AtCoder Beginner Contest 100 C(思维)

*3 or /2
题目大意:有n个数,每次操作将第i个数*3或/2,得到结果必须为整数,且每次操作必须有一次为/2,求最大操作次数.
一开始看很懵比,感觉肯定是思维题,对着样例猜了个结论竟然就过了大数据。。。
思路:奇数只能 * 3,所以只考虑偶数.对于一个偶数,可以分解成2^n * a,显然a为奇数,那么如果这个偶数乘3为2^n * 3a,3a也显然为奇数,所以证得:任意一个偶数无论乘多少次3它能/2的次数仍然为n不变.
由于次数最大,所以每次只让一个偶数/2。
所以答案就是统计每个偶数为2^n * a中n的和。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <queue>
 6 #define ll long long
 7 #define out(a) printf("%d",a)
 8 using namespace std;
 9 int n,cnt;
10 int a[100050];
11 int read()
12 {
13     int s=0,t=1; char c;
14     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
15     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
16     return s*t;
17 }
18 int main()
19 {
20     n=read();
21     for (int i=1;i<=n;i++)
22       a[i]=read();
23     for (int i=1;i<=n;i++)
24       if (a[i]%2==0) {
25         while (true) {
26           if (a[i]%2==0) cnt++,a[i]/=2;
27           else break;
28         }
29     }
30     out(cnt);
31     return 0;
32 }

原文地址:https://www.cnblogs.com/Kaleidoscope233/p/9277274.html

时间: 2024-11-08 08:32:26

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