Foreigntty的方程math [数学]

Posted 2020-10-16 BearChild

tty的方程math

Time Limit: 50 Sec  Memory Limit: 128 MB

Description　

　　给定n、m、k、p。

　　a+b+c=n, a^2+b^2+c^2=m, a^3+b^3+c^3=k。

　　求a^p+b^p+c^p。

Input

　　输入n、m、k、p

Output

　　以A/B形式表示答案。

Sample Input

　　5 7 11 4

Sample Output

　　27/1

HINT

　　0<=n,m,k <=20，0<=p<=10

Solution

 1 \\begin {align}
 2 &a^p+b^p+c^p\\:\\:①
 3 \\\\
 4 \\\\&(a+b+c)*(a^{p-1}+b^{p-1}+c^{p-1})\\:\\:②
 5 \\\\=&a^{p}+b^{p}+c^{p}+ab^{p-1}+ac^{p-1}+ba^{p-1}+bc^{p-1}+ca^{p-1}+cb^{p-1}
 6 \\\\
 7 \\\\∴&①-②
 8 \\\\=&-(ab^{p-1}+ac^{p-1}+ba^{p-1}+bc^{p-1}+ca^{p-1}+cb^{p-1})
 9 \\\\
10 \\\\&(ab+bc+ac)*(a^{p-2}+b^{p-2}+c^{p-2})\\:\\:③
11 \\\\=&ba^{p-1}+ab^{p-1}+abc^{p-2}+bca^{p-2}+cb^{p-1}+bc^{p-1}+ca^{p-1}+acb^{p-1}+ac^{p-1}
12 \\\\
13 \\\\∴&①-②+③=
14 \\\\=&bca^{p-2}+acb^{p-2}+abc^{p-2}
15 \\\\=&abc(a^{p-3}+b^{p-3}+c^{p-3})\\:\\:④
16 \\\\
17 \\\\∴&①-②+③=④即①=②-③+④
18 \\\\
19 \\\\&那么现在问题就是如何求出(a+b+c),(ab+ac+bc),(abc)。
20 \\\\&首先题目给定了a+b+c=n,a^2+b^2+c^2=m,a^3+b^3+c^3=k，那么：
21 \\\\
22 \\\\1.&a+b+c=n
23 \\\\2.&(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc
24 \\\\&ab+ac+bc=\\frac{n^2-m}2
25 \\\\3.&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc.
26 \\\\&a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=(ab+ac+bc)*(a+b+c)-3abc
27 \\\\&n^3=k+3* \\frac{(n^2-m) * n}2-9abc+6abc
28 \\\\&abc=\\frac{2*k+3n*(n^2-m)-2*n^3}{6}
29 \\end {align}

math

　　

Code

 1 #include<iostream>
 2 #include<string>
 3 #include<algorithm>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cmath>
 8 using namespace std;
 9 typedef long long s64;
10 
11 const int ONE = 1000005;
12 const int MOD = 1e9 + 7;
13 
14 int get()
15 {
16         int res = 1, Q = 1; char c;
17         while( (c = getchar()) < 48 || c > 57)
18             if(c == \'-\') Q = -1;
19         if(Q) res = c - 48;
20         while( (c = getchar()) >= 48 && c <= 57)
21             res = res * 10 + c - 48;
22         return res * Q;
23 }
24 
25 s64 gcd(s64 a, s64 b)
26 {
27         while(s64 r = a % b) {a = b; b = r;}
28         return b;
29 }
30 
31 int n, m, k, p;
32 struct power
33 {
34         s64 fz, fm;
35 }Ans[ONE], A, B, C, now;
36 
37 power Deal(power a, power b)
38 {
39         s64 fz = a.fz * b.fm + b.fz * a.fm, fm = a.fm * b.fm;
40         int p1 = fz > 0, p2 = fm > 0;
41         fz = abs(fz), fm = abs(fm);
42         s64 r = gcd(fz, fm);
43         return (power){p1 * fz / r, p2 * fm / r};
44 }
45 
46 int main()
47 {
48         cin >> n >> m >> k >> p;
49         if(p == 0) {printf("3"); return 0;}
50         Ans[1] = (power){n, 1};
51         Ans[2] = (power){m, 1};
52         Ans[3] = (power){k, 1};
53 
54         A = (power){n, 1};
55         B = (power){n * n - m, 2};
56         C = (power){2 * k + 3 * n * (n * n - m) - 2 * n * n * n, 6};
57 
58         for(int i = 4; i <= p; i++)
59         {
60             power now = (power){A.fz * Ans[i-1].fz, A.fm * Ans[i-1].fm};
61             now = Deal(now, (power){C.fz * Ans[i-3].fz, C.fm * Ans[i-3].fm});
62             now = Deal(now, (power){-B.fz * Ans[i-2].fz, B.fm * Ans[i-2].fm});
63 
64             Ans[i] = now;
65         }
66 
67         printf("%d/%d", Ans[p].fz, Ans[p].fm);
68 }

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