2020劳动节天天乐（4）题解

A - 两只脑斧

题意：给你n个字符串，根据图判断呼还是吸还是停顿

思路：模拟，细心发现，偶数和7的时候是吸，0是停顿，其余是呼。

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 ll a[N],dp[N][2];//0为正，1为负
15 int main()
16 {
17     ll i, j, k;
18     ll n, m, t;
19     ll count1 = 0, count2 = 0;
20     cin >> t;
21     while (t--)
22     {
23         string s;
24         cin >> s;
25         if (s[0] - '0' == 0)
26             cout << "X";
27         else if (s[0] - '0' == 7)
28             cout << "I";
29         else if ((s[0] - '0') % 2 == 0)
30             cout << "I";
31         else
32             cout << "E";
33     }
34 
35 }

C - 赛尔逵传说

题意：零氪有k点攻击力，c个“力量炖水果”（吃一个加k点攻击，可多次吃），给你n个怪di血xi攻击，问零氪最少承受攻击力。

思路：贪心，零氪打死一只怪要（d/k）-1次，而力量水果c其实实际上是减少零氪打死怪的攻击次数，怎样减最好呢？当然减去攻击力最高，先排除零氪能一下秒的怪，再排序即可

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 const int MOD = 998244353;
15 struct node {
16     ll d, x;
17 }a[N];
18 bool cmp(const node& a, const node& b)
19 {
20     if (a.x == b.x)return a.d < b.d;
21     else
22         return a.x > b.x;
23 }
24 int main()
25 {
26     ll i, j, k;
27     ll n, m, t,c;
28     ll ret = 0, count1 = 0, count2 = 0;
29     cin >> n >> k>>c;
30     for (i = 1; i <= n; i++)
31     {
32         ll d, x;
33         cin >> d >> x;
34         if (d > k)
35         {
36             a[ret].d = d, a[ret].x = x;
37             ret++;
38         }
39     }
40     sort(a, a + ret, cmp);
41     ll count = 0;//共计次数
42     ll ans = 0;//受伤量
43     for (i = 0; i < ret; i++)
44     {    
45         if (a[i].d % k != 0)
46             count = a[i].d / k;
47         else
48             count = a[i].d / k - 1;
49         if (c != 0)//优先慢慢用完
50         {
51             if (count > c)
52             {
53                 count -= c;
54                 c = 0;
55             }
56             else
57             {
58                 c -= count;
59                 count = 0;
60             }
61         }
62         ans += count * a[i].x;
63     }
64     cout << ans << endl;
65 
66 }

D - 只有一端开口的瓶子

题意：给你一个一个乱序的全排列，可以有k个栈存入，有push，pop，move操作，要使序列变成递增序列，问最小k是多少

思路：其实仔细想想你最多只需要俩个栈就可以，跟汉诺塔的原理差不多的，因为是全排列，你可以拿一个栈模拟能否成功，能就一个，不能就两个

呈上代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 const int MOD = 998244353;
15 ll a[N];
16 stack<ll>b;
17 int main()
18 {
19     ll i, j, k;
20     ll n, m, t,c;
21     cin >> t;
22     while (t--)
23     {
24         while (!b.empty())
25             b.pop();
26         cin >> n;
27         for (i = 1; i <= n; i++)
28             cin >> a[i];
29         ll pos = 1;//找到pos就可以直接pop出来
30         for (i = 1; i <= n; i++)//
31         {
32             b.push(a[i]);
33             while (!b.empty()&&b.top() == pos)
34             {
35                 pos++;
36                 b.pop();
37             }
38         }
39         if (b.empty())
40             cout << 1 << endl;
41         else
42             cout << 2 << endl;
43     }
44 
45 }

E - 风王之瞳

题意：给N*M的网格，问你能有多少个正方形

思路：纯属规律题，也有dp的做法，就不详细介绍dp做法了。

我在其他地方找到的，懒得画了

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 ll a[N],dp[N][2];//0为正，1为负
15 int main()
16 {
17     ll i, j, k;
18     ll n, m, t;
19     ll count1 = 0, count2 = 0;
20     cin >> t;
21     while (t--)
22     {
23         cin >> n >> m;
24         ll sum = 0;
25         for (i = 1; i <= min(n, m); i++)
26             sum += (m - i + 1) * (n - i + 1)*i;
27         cout << sum << endl;
28     }
29 
30 }

G - 目标是成为数论大师

题意：函数 $f (x) = \sqrt{a x} + b$

思路：把 f(x)=x代入得x^2-(2*b+a)*x+b^2=0,求解二元一次方程根就好，注意要排除增根

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 int main()
15 {
16     ll i, j, k;
17     ll n, m, t;
18     ll count1 = 0, count2 = 0;
19     cin >> t;
20     while (t--)
21     {
22         ll a, b;
23         ll x1, x2;
24         cin >> a >> b;
25         ll delta = 4 * a * b + a * a;
26         if (delta > 0)//两个点
27         {
28             
29             x1 = ((2 * b + a) + sqrt(delta))/2;
30             x2 =( (2 * b + a) - sqrt(delta)) / 2 ;
31             ll flag1 = 0, flag2 = 0;
32             //去增根
33             if (sqrt(a * x1) + b == x1)
34                 flag1 = 1;
35             if (sqrt(a * x2) + b == x2)
36                 flag2 = 1;
37             if(flag1&&flag2)
38             {
39                 cout << 2 << endl;
40                 cout << x2 << " " << x1 << endl;
41             }
42             else
43             {
44                 cout << 1 << endl;
45                 if (flag1)
46                     cout << x1 << endl;
47                 else
48                     cout << x2 << endl;
49             }
50         }
51         else
52         {
53             cout << 1 << endl;
54             cout << (2 * b + a)/2 << endl;
55         }
56     }
57 
58 }

H - 金色传说

题意：在计算机必输入n位，可输0-9 + -十二个键，但要符合计算式要求，输够n位后自动生成解，问所有解的和是多少

思路：没什么思路，靠的数学感觉加上计算器的辅助，网上有dp的做法

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 const int N = 5e5 + 5;
12 const int mod = 1e9 + 9;
13 const int MOD = 998244353;
14 ll a[N];
15 ll fastPow(ll a, ll n)
16 {
17     ll base = a%MOD;
18     ll res = 1;
19     while (n)
20     {
21         if (n & 1)
22             res = (res*base)%MOD;
23         base = (base*base)%MOD;
24         n >>= 1;
25     }
26     return res % MOD;
27 }
28 int main()
29 {
30     ll i, j, k;
31     ll n, m, t,c;
32     a[1] = 45, a[2] = 4950;
33     ll sum = 0;
34     for (i = 3; i <=500001; i++)
35     {
36         ll x = fastPow(10, i);
37         sum =( sum + a[i - 2] * 20)%MOD;
38         a[i] = ((x * (x - 1) / 2)%MOD  + sum)%MOD;
39         sum = (sum*10)%MOD;
40     }
41     cin >> t;
42     while (t--)
43     {
44         cin >> n;
45         //cout << "TEXT：";
46         //ll x = fastPow(10, n);
47         //cout << x * (x-1) / 2 << endl;
48         cout << a[n]%MOD << endl;
49     }
50 
51 }

I - 多项式求导

题意：给你最高n次项多项式，求k次求导后n次项各个系数

思路：根据求导方式来求，求到的结果再把原来数组更新就好，数据不大

代码

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 #include<map>
 5 #include<stack>
 6 #include<cstring>
 7 #include<cmath>
 8 using namespace std;
 9 #define ll long long
10 #define double long double
11 #define eps 0.0000000001//偏差值1e8
12 const int N = 2e5 + 5;
13 const int mod = 1e9 + 9;
14 const int MOD = 998244353;
15 ll a[N];
16 ll b[N];
17 int main()
18 {
19     ll i, j, k;
20     ll n, m, t;
21     ll count1 = 0, count2 = 0;
22     cin >> n >> k;
23     for (i = 0; i <= n; i++)
24         cin >> a[i];
25     b[0] = 0;
26     for (j = 0; j < k; j++)
27     {
28         for (i = 0; i < n; i++)
29         {
30             b[i + 1] = (n - i) * a[i]%MOD;
31         }
32         for (i = 0; i <= n; i++)
33             a[i] = b[i];
34     }
35     for (i = 0; i <= n; i++)
36         cout << b[i] << " ";
37 
38 }

posted @ 2020-05-05 17:53 ch_hui 阅读(208) 评论(0) 编辑收藏举报

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