【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}

　　

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 = 1000005;
const int MOD = 1e9 + 7;

int get()
{
        int res = 1, Q = 1; char c;
        while( (c = getchar()) < 48 || c > 57)
            if(c == '-') Q = -1;
        if(Q) res = c - 48;
        while( (c = getchar()) >= 48 && c <= 57)
            res = res * 10 + c - 48;
        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 > 0, p2 = fm > 0;
        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 == 0) {printf("3"); return 0;}
        Ans[1] = (power){n, 1};
        Ans[2] = (power){m, 1};
        Ans[3] = (power){k, 1};

        A = (power){n, 1};
        B = (power){n * n - m, 2};
        C = (power){2 * k + 3 * n * (n * n - m) - 2 * n * n * n, 6};

        for(int i = 4; i <= p; i++)
        {
            power now = (power){A.fz * Ans[i-1].fz, A.fm * Ans[i-1].fm};
            now = Deal(now, (power){C.fz * Ans[i-3].fz, C.fm * Ans[i-3].fm});
            now = Deal(now, (power){-B.fz * Ans[i-2].fz, B.fm * Ans[i-2].fm});

            Ans[i] = now;
        }

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