题解：CF1688B Patchouli's Magical Talisman
对于一个奇数，不需要进行操作；对于一个偶数，只要有奇数存在，显然直接与奇数合并最优。因此，若序列中存在奇数，则由奇数+偶数=奇数的性质，可以将所有的偶数与其合并；若不存在奇数，则找到处理最少次数后能变为奇数的偶数，进行操作，然后将剩余的偶数与其合并。
代码如下：
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| #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define init(x) memset (x,0,sizeof (x))
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 2e5 + 5;
const int MOD = 1e9 + 7;
inline int read ();
int t,n,mx,cnt,a[MAX];
int main ()
{
t = read ();
while (t--)
{
n = read ();cnt = 0;mx = INF;
for (int i = 1;i <= n;++i) a[i] = read ();
for (int i = 1;i <= n;++i)
if (a[i] & 1) ++cnt;
for (int i = 1;i <= n;++i)
{
if (a[i] & 1) continue;
int p = 0;
while (a[i] % 2 == 0)
{
a[i] /= 2;
++p;
}
mx = min (mx,p);
}
if (cnt == n) printf ("0\n");
else if (!cnt) printf ("%d\n",n - 1 + mx);
else printf ("%d\n",n - cnt);
}
return 0;
}
inline int read ()
{
int s = 0;int f = 1;
char ch = getchar ();
while ((ch < '0' || ch > '9') && ch != EOF)
{
if (ch == '-') f = -1;
ch = getchar ();
}
while (ch >= '0' && ch <= '9')
{
s = s * 10 + ch - '0';
ch = getchar ();
}
return s * f;
}
|
