【Foreign】tty的方程math [数学]

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

 \begin {align}
 &a^p+b^p+c^p\:\:①
 \\
 \\&(a+b+c)*(a^{p-1}+b^{p-1}+c^{p-1})\:\:②
 \\=&a^{p}+b^{p}+c^{p}+ab^{p-1}+ac^{p-1}+ba^{p-1}+bc^{p-1}+ca^{p-1}+cb^{p-1}
 \\
 \\∴&①-②
 \\=&-(ab^{p-1}+ac^{p-1}+ba^{p-1}+bc^{p-1}+ca^{p-1}+cb^{p-1})
 \\
 \\&(ab+bc+ac)*(a^{p-2}+b^{p-2}+c^{p-2})\:\:③
 \\=&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}
 \\
 \\∴&①-②+③=
 \\=&bca^{p-2}+acb^{p-2}+abc^{p-2}
 \\=&abc(a^{p-3}+b^{p-3}+c^{p-3})\:\:④
 \\
 \\∴&①-②+③=④即①=②-③+④
 \\
 \\&那么现在问题就是如何求出(a+b+c),(ab+ac+bc),(abc)。
 \\&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：
 \\
 \\1.&a+b+c=n
 \\2.&(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc
 \\&ab+ac+bc=\frac{n^2-m}2
 \\3.&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc.
 \\&a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=(ab+ac+bc)*(a+b+c)-3abc
 \\&n^3=k+3* \frac{(n^2-m) * n}2-9abc+6abc
 \\&abc=\frac{2*k+3n*(n^2-m)-2*n^3}{6}
 \end {align}

math

　　

Code

 #include<iostream>
 #include<string>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 using namespace std;
 typedef long long s64;

 const int ONE = ;
 const int MOD = 1e9 + ;

 int get()
 {
         int res = , Q = ; char c;
         while( (c = getchar()) <  || c > )
             if(c == '-') Q = -;
         if(Q) res = c - ;
         while( (c = getchar()) >=  && c <= )
             res = res *  + c - ;
         return res * Q;
 }

 s64 gcd(s64 a, s64 b)
 {
         while(s64 r = a % b) {a = b; b = r;}
         return b;
 }

 int n, m, k, p;
 struct power
 {
         s64 fz, fm;
 }Ans[ONE], A, B, C, now;

 power Deal(power a, power b)
 {
         s64 fz = a.fz * b.fm + b.fz * a.fm, fm = a.fm * b.fm;
         int p1 = fz > , p2 = fm > ;
         fz = abs(fz), fm = abs(fm);
         s64 r = gcd(fz, fm);
         return (power){p1 * fz / r, p2 * fm / r};
 }

 int main()
 {
         cin >> n >> m >> k >> p;
         if(p == ) {printf(""); return ;}
         Ans[] = (power){n, };
         Ans[] = (power){m, };
         Ans[] = (power){k, };

         A = (power){n, };
         B = (power){n * n - m, };
         C = (power){ * k +  * n * (n * n - m) -  * n * n * n, };

         for(int i = ; i <= p; i++)
         {
             power now = (power){A.fz * Ans[i-].fz, A.fm * Ans[i-].fm};
             now = Deal(now, (power){C.fz * Ans[i-].fz, C.fm * Ans[i-].fm});
             now = Deal(now, (power){-B.fz * Ans[i-].fz, B.fm * Ans[i-].fm});

             Ans[i] = now;
         }

         printf("%d/%d", Ans[p].fz, Ans[p].fm);
 }