Atcoder CADDi 2018 Solution

C - Product and GCD

Solved.

题意：

给出$n个数$的乘积，求$这n个数$的最大的可能是GCD

思路：

分解质因子，那么$每个质因子的贡献就是其质因子个数/ n的乘积$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 #define ll long long
 5 ll n, p;
 6
 7 int main()
 8 {
 9     while (scanf("%lld%lld", &n, &p) != EOF)
10     {
11         if (n == 1)
12         {
13             printf("%lld\n", p);
14             continue;
15         }
16         ll res = 1;
17         for (ll i = 2; i * i <= p; ++i)
18         {
19             ll tmp = 0;
20             while (p % i == 0)
21             {
22                 ++tmp;
23                 p /= i;
24             }
25             while (tmp >= n)
26             {
27                 tmp -= n;
28                 res *= i;
29             }
30         }
31         printf("%lld\n", res);
32     }
33     return 0;
34 }

View Code

D - Harlequin

Solved.

题意：

有$n种颜色的苹果，每种颜色有a_i个，每次可以选择若干个不同颜色的苹果拿掉，轮流拿，谁不能拿谁输$

问先手胜还是后手胜

思路：

必败局面是当前场上所有颜色的苹果都是偶数个，这样的话，你拿什么，对方跟着你拿，对方肯定胜利

那么必胜局面就是当前场上存在有若干种奇数个颜色的苹果，取掉这些苹果，转换成必败局面留给对方就好了

简单判一判就没了

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 int n, x;
 5
 6 int main()
 7 {
 8     while (scanf("%d", &n) != EOF)
 9     {
10         x = 0;
11         for (int i = 1, y; i <= n; ++i)
12         {
13             scanf("%d", &y);
14             x += y & 1;
15         }
16         puts(x ? "first" : "second");
17     }
18     return 0;
19 }

View Code

E - Negative Doubling

Upsolved.

题意：

给出n个数，对每个数的操作只有 $\cdot -2$ ，求最少多少次操作使得序列变成非下降序列

思路：

我们考虑序列中如果存在正数和负数，那么必然存在一个界限，左边全是负数，右边全是正数

那么我们可以预处理出

$f[i] 表示使得从i开始是一个非下降序列的最小花费$

$g[i]表示 i以及i以前是一个非下降序列，且全都是负数的最小花费$

维护的过程可以用一个队列

刚开始以为如果当前数比之前的数不满足大小关系，后面的数或者前面的数全都要动

但实际上并不是这样，

比如说

5 3 4 1000000

推到5的时候只有 3 和 4 要改变

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3
  4 #define ll long long
  5 #define N 200010
  6 #define pii pair <int, int>
  7 int n;
  8 ll a[N], b[N], f[N], g[N], lazy;
  9
 10 void Run()
 11 {
 12     while (scanf("%d", &n) != EOF)
 13     {
 14         for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
 15         queue <pii> q;
 16         f[n] = 0;
 17         for (int i = n - 1; i >= 1; --i)
 18         {
 19             f[i] = f[i + 1];
 20             if (a[i] > a[i + 1])
 21             {
 22                 ll tmp = a[i + 1];
 23                 int cnt = 0;
 24                 while (tmp < a[i])
 25                 {
 26                     tmp *= 4;
 27                     cnt += 2;
 28                 }
 29                 int x = i;
 30                 while (!q.empty())
 31                 {
 32                     f[i] += 1ll * (q.front().second - x) * cnt;
 33                     if (q.front().first <= cnt) cnt -= q.front().first;
 34                     else
 35                     {
 36                         pii tmpp = q.front();
 37                         tmpp.first -= cnt;
 38                         cnt = 0;
 39                         q.pop();
 40                         q.push(tmpp);
 41                         break;
 42                     }
 43                     x = q.front().second;
 44                     q.pop();
 45                 }
 46                 f[i] += 1ll * (n - x) * cnt;
 47             }
 48             else
 49             {
 50                 ll tmp = a[i];
 51                 int cnt = 0;
 52                 while (tmp * 4 <= a[i + 1])
 53                 {
 54                     tmp *= 4;
 55                     cnt += 2;
 56                 }
 57                 if (cnt) q.push(pii(cnt, i));
 58             }
 59         }
 60         while (!q.empty()) q.pop();
 61         g[1] = 1;
 62         b[1] = -a[1];
 63         for (int i = 2; i <= n; ++i)
 64         {
 65             g[i] = g[i - 1] + 1;
 66             b[i] = -a[i];
 67             if (b[i] < b[i - 1])
 68             {
 69                 ll tmp = b[i - 1];
 70                 int cnt = 0;
 71                 while (tmp > b[i])
 72                 {
 73                     tmp *= 4;
 74                     cnt += 2;
 75                 }
 76                 int x = i;
 77                 while (!q.empty())
 78                 {
 79                     g[i] += 1ll * (x - q.front().second) * cnt;
 80                     if (q.front().first <= cnt) cnt -= q.front().first;
 81                     else
 82                     {
 83                         pii tmpp = q.front();
 84                         q.pop();
 85                         tmpp.first -= cnt;
 86                         cnt = 0;
 87                         q.push(tmpp);
 88                         break;
 89                     }
 90                     x = q.front().second;
 91                     q.pop();
 92                 }
 93                 g[i] += 1ll * (x - 1) * cnt;
 94             }
 95             else
 96             {
 97                 ll tmp = b[i];
 98                 int cnt = 0;
 99                 while (tmp * 4 >= b[i - 1])
100                 {
101                     tmp *= 4;
102                     cnt += 2;
103                 }
104                 if (cnt) q.push(pii(cnt, i));
105             }
106         }
107         ll res = min(f[1], g[n]);
108         for (int i = 1; i < n; ++i) res = min(res, g[i] + f[i + 1]);
109         printf("%lld\n", res);
110     }
111 }
112
113
114 int main()
115 {
116     #ifdef LOCAL
117         freopen("Test.in", "r", stdin);
118     #endif
119
120     Run();
121     return 0;
122 }

View Code

转载于:https://www.cnblogs.com/Dup4/p/10163562.html

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