CF 1368B Codeforces Subsequences

传送门

题目：给定一个串“codeforces”，给定一个n，让你在原串上任意位置添加任意个数的字符构造出一个字符串s，使得至少有n组子序列能够组成“codeforces”，需要构造出的串长度最短，再输出构造出的s

思路：根据排列组合的性质，我们尽量平均“codeforces”上每位的个数即可。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8 #include <stack>
 9 #include <map>
10  
11 using namespace std;
12   
13 #define ll long long
14 #define pb push_back
15 #define fi first
16 #define se second
17   
18 const int N = 2e6 + 10;
19 
20 
21 void solve()
22 {
23     ll n;
24     scanf("%lld", &n);
25     ll p = 1;
26     vector<ll > Pow;
27     while(1){
28         ll k = p*p*p*p*p*p*p*p*p*p;
29         Pow.pb(k);
30         p++;
31         if(k > 1e16) break;
32     }
33     int _size = Pow.size();
34     int inx = -1;
35     for(int i = 0; i < _size; ++i){
36         if(Pow[i] > n) break;
37         //找到满足 i^10 <= n < (i+1)^10 的i
38         inx = i;
39     }
40     // cout << "inx = " << inx << endl;
41     // cout << "pow = " << Pow[inx] << endl;
42     inx++;
43 
44     //下面确定几个字符需要比平均多一个
45     //----------------------------
46     ll a[20], b[20];
47     a[0] = 1;
48     a[1] = inx;
49     for(int i = 2; i <= 10; ++i) a[i] = a[i - 1] * inx;
50     b[0] = 1;
51     b[1] = inx + 1;
52     for(int i = 2; i <= 10; ++i) b[i] = b[i - 1] * (inx + 1);
53     ll c[20];
54     for(int i = 0; i <= 10; ++i){
55         c[i] = b[i] * a[10 - i];
56     }
57 
58     int cnt = 0;
59     for(int i = 1; i <= 10; ++i){
60         if(n > c[i - 1] && n <= c[i]) cnt = i;
61     }
62     //-----------------------------
63 
64     //if(n == Pow[inx - 1]) cnt = 0;
65     for(int i = 1; i <= inx; ++i) putchar('c');
66     if(cnt-- > 0) putchar('c');
67     for(int i = 1; i <= inx; ++i) putchar('o');
68     if(cnt-- > 0) putchar('o');
69     for(int i = 1; i <= inx; ++i) putchar('d');
70     if(cnt-- > 0) putchar('d');
71     for(int i = 1; i <= inx; ++i) putchar('e');
72     if(cnt-- > 0) putchar('e');
73     for(int i = 1; i <= inx; ++i) putchar('f');
74     if(cnt-- > 0) putchar('f');
75     for(int i = 1; i <= inx; ++i) putchar('o');
76     if(cnt-- > 0) putchar('o');
77     for(int i = 1; i <= inx; ++i) putchar('r');
78     if(cnt-- > 0) putchar('r');
79     for(int i = 1; i <= inx; ++i) putchar('c');
80     if(cnt-- > 0) putchar('c');
81     for(int i = 1; i <= inx; ++i) putchar('e');
82     if(cnt-- > 0) putchar('e');
83     for(int i = 1; i <= inx; ++i) putchar('s');
84     if(cnt-- > 0) putchar('s');
85     puts("");
86 }
87   
88 int main()
89 {
90     // ios::sync_with_stdio(false);
91     // cin.tie(0);
92     // cout.tie(0);
93 
94     solve();
95  
96     return 0;
97 }

 

posted @ 2020-07-28 23:47  SummerMingQAQ  阅读(244)  评论(0)  编辑  收藏  举报