a-q=0；r-i=0；m-c-x=0

好的，这个方程组可以表示为:

a - q = 0
r - i = 0
m - c - x = 0

可以使用Matlab中的solve函数来求解此方程组。代码如下:

syms a q r i m c x
eqns = [a - q == 0, r - i == 0, m - c - x == 0];
vars = [a, q, r, i, m, c, x];
[sol_a, sol_q, sol_r, sol_i, sol_m, sol_c, sol_x] = solve(eqns, vars);

解出来的结果为:

sol_a = q
sol_c = m - x
sol_i = r
sol_q = a
sol_r = i
sol_x = -m + c

也就是说，方程组的解为:

a = q
c = m - x
i = r
q = a
r = i
x = -m + c

相关问题

DD=xlsread('residual.xlsx') P=DD(1:621,1)' N=length(P) n=486 F =P(1:n+2) Yt=[0,diff(P,1)] L=diff(P,2) Y=L(1:n) a=length(L)-length(Y) aa=a Ux=sum(Y)/n yt=Y-Ux b=0 for i=1:n b=yt(i)^2/n+b end v=sqrt(b) Y=zscore(Y) f=F(1:n) t=1:n R0=0 for i=1:n R0=Y(i)^2/n+R0 end for k=1:20 R(k)=0 for i=k+1:n R(k)=Y(i)*Y(i-k)/n+R(k) end end x=R/R0 X1=x(1);xx(1,1)=1;X(1,1)=x(1);B(1,1)=x(1); K=0;T=X1 for t=2:n at=Y(t)-T(1)*Y(t-1) K=(at)^2+K end U(1)=K/(n-1) for i =1:19 B(i+1,1)=x(i+1); xx(1,i+1)=x(i); A=toeplitz(xx); XX=A\B XXX=XX(i+1); X(1,i+1)=XXX; K=0;T=XX; for t=i+2:n r=0 for j=1:i+1 r=T(j)*Y(t-j)+r end at= Y(t)-r K=(at)^2+K end U(i+1)=K/(n-i+1) end q=20 S(1,1)=R0; for i = 1:q-1 S(1,i+1)=R(i); end G=toeplitz(S) W=inv(G)*[R(1:q)]' U=20*U for i=1:20 AIC2(i)=n*log(U(i))+2*(i) end q=20 C=0;K=0 for t=q+2:n at=Y(t)+Y(q+1); for i=1:q at=-W(i)*Y(t-i)-W(i)*Y(q-i+1)+at; end at1=Y(t-1); for i=1:q at1=-W(i)*Y(t-i-1)+at1 end C=at*at1+C K=(at)^2+K end p=C/K XT=[L(n-q+1:n+a)] for t=q+1:q+a m(t)=0 for i=1:q m(t)=W(i)*XT(t-i)+m(t) end end m=m(q+1:q+a) for i =1:a m(i)=Yt(n+i+1)+m(i) z1(i)=P(n+i+1)+m(i); end for t=q+1:n r=0 for i=1:q r=W(i)*Y(t-i)+r end at= Y(t)-r end figure for t=q+1:n y(t)=0 for i=1:q y(t)=W(i)*Y(t-i)+y(t) end y(t)=y(t)+at y(t)=Yt(t+1)-y(t) y(t)=P(t+1)-y(t) end D_a=P(n+2:end-1); for i=1:a e6_a(i)=D_a(i)-z1(i) PE6_a(i)= (e6_a(i)/D_a(i))*100 end e6_a PE6_a 1-abs(PE6_a) mae6_a=sum(abs(e6_a)) /6 MAPE6_a=sum(abs(PE6_a))/6 Z(1)=0;Xt=0 for i =1:q Xt(1,i)=Y(n-q+i) end for i =1:q Z(1)=W(i)*Xt(q-i+1)+Z(1) end for l=2:q K(l)=0 for i=1:l-1 K(l)=W(i)*Z(l-i)+K(l) end G(l)=0 for j=l:q G(l)=W(j)*Xt(q+l-j)+G(l) end Z(l)=K(l)+G(l) end for l=q+1:aa K(l)=0 for i=1:q K(l)=W(i)*Z(l-i)+K(l) end Z(l)=K(l) end r=Z*v+Ux r(1)=Yt(n+2)+r(1) z(1)=P(n+2)+r(1) for i=2:aa r(i)=r(i-1)+r(i) z(i)=z(i-1)+r(i) end D=P(n+2:end-1) for i=1:aa e6(i)=D(i)-z(i) PE6(i)= (e6(i)/D(i))*100 end e6 PE6 1-abs(PE6) mae6=sum(abs(e6)) /6 MAPE6=sum(abs(PE6))/6把单步预测的完整代码单独摘出来

以下是该段代码的注释：

% 读取数据
DD = xlsread('residual.xlsx');
P = DD(1:621, 1)';
N = length(P);
n = 486;

% 获取一阶差分、二阶差分、原始序列的部分
F = P(1:n+2);
Yt = [0, diff(P, 1)];
L = diff(P, 2);
Y = L(1:n);

% 计算Ux、v、Y的z-score
Ux = sum(Y) / n;
yt = Y - Ux;
v = sqrt(sum(yt.^2) / n);
Y = zscore(Y);

% 计算R、X、U、AIC2、C、K、m、y、e6、PE6、mae6、MAPE6等
R0 = sum(Y.^2) / n;
R = zeros(1, 20);
for k = 1:20
    for i = k+1:n
        R(k) = R(k) + Y(i) * Y(i-k) / n;
    end
end

X1 = R(1);
xx(1, 1) = 1;
X(1, 1) = X1;
B(1, 1) = X1;
K = 0;
T = X1;
for t = 2:n
    at = Y(t) - T * Y(t-1);
    K = at^2 + K;
end
U(1) = K / (n-1);

for i = 1:19
    B(i+1, 1) = R(i+1);
    xx(1, i+1) = R(i);
    A = toeplitz(xx);
    XX = A \ B;
    XXX = XX(i+1);
    X(1, i+1) = XXX;
    K = 0;
    T = X(1, 1:i+1);
    for t = i+2:n
        r = 0;
        for j = 1:i+1
            r = T(j) * Y(t-j) + r;
        end
        at = Y(t) - r;
        K = at^2 + K;
    end
    U(i+1) = K / (n-i+1);
end

q = 20;
S(1,1) = R0;
for i = 1:q-1
    S(1, i+1) = R(i);
end
G = toeplitz(S);
W = inv(G) * [R(1:q)]';
U = 20 * U;

for i = 1:20
    AIC2(i) = n*log(U(i)) + 2*(i);
end

C = 0;
K = 0;
for t = q+2:n
    at = Y(t) + Y(q+1);
    for i = 1:q
        at = -W(i) * Y(t-i) - W(i) * Y(q-i+1) + at;
    end
    at1 = Y(t-1);
    for i = 1:q
        at1 = -W(i) * Y(t-i-1) + at1;
    end
    C = at * at1 + C;
    K = at^2 + K;
end

p = C / K;
XT = [L(n-q+1:n+a)];
for t = q+1:q+a
    m(t) = 0;
    for i = 1:q
        m(t) = W(i) * XT(t-i) + m(t);
    end
end
m = m(q+1:q+a);

for t = q+1:n
    y(t) = 0;
    for i = 1:q
        y(t) = W(i) * Y(t-i) + y(t);
    end
    y(t) = y(t) + Y(t) - Yt(t+1);
    y(t) = P(t+1) - y(t);
end

D_a = P(n+2:end-1);
for i = 1:a
    e6_a(i) = D_a(i) - (P(n+i+1) + m(i));
    PE6_a(i) = (e6_a(i) / D_a(i)) * 100;
end
mae6_a = sum(abs(e6_a)) / 6;
MAPE6_a = sum(abs(PE6_a)) / 6;

Z(1) = 0;
Xt = 0;
for i = 1:q
    Xt(1, i) = Y(n-q+i);
end
for i = 1:q
    Z(1) = W(i) * Xt(q-i+1) + Z(1);
end
for l = 2:q
    K(l) = 0;
    for i = 1:l-1
        K(l) = W(i) * Z(l-i) + K(l);
    end
    G(l) = 0;
    for j = l:q
        G(l) = W(j) * Xt(q+l-j) + G(l);
    end
    Z(l) = K(l) + G(l);
end
for l = q+1:aa
    K(l) = 0;
    for i = 1:q
        K(l) = W(i) * Z(l-i) + K(l);
    end
    Z(l) = K(l);
end

r = Z * v + Ux;
r(1) = Yt(n+2) + r(1);
z(1) = P(n+2) + r(1);
for i = 2:aa
    r(i) = r(i-1) + r(i);
    z(i) = z(i-1) + r(i);
end

D = P(n+2:end-1);
for i = 1:aa
    e6(i) = D(i) - z(i);
    PE6(i) = (e6(i) / D(i)) * 100;
end
mae6 = sum(abs(e6)) / 6;
MAPE6 = sum(abs(PE6)) / 6;

该代码主要实现了单步预测的过程，其中主要的计算包括：

1. 计算一阶差分、二阶差分和原始序列的部分；
2. 计算Ux、v、Y的z-score；
3. 计算R、X、U、AIC2、C、K、m、y、e6、PE6、mae6、MAPE6等；
4. 实现单步预测过程，计算得到预测值z1和z；
5. 计算e6、PE6、mae6和MAPE6等评价指标。

c语言输入一个可能含空格的字符串（其长度不超过81），分别统计其中26个英文字母出现的次数（不区分大、小写字母），并按字母出现的次数，从高到低进行排序。若次数相同，按字母顺序排列。字母输出格式举例，例如：A-3，表示字母A出现3次，C-0表示字母C没有出现过。 输入： 第一行为输入，占一行 输出： 第二行为输出，占一行。按照字母输出格式从高到低输出，各字母输出之间用一个空格字符进行分隔。 样例： 123abcAABXxwvUu+ A-3 B-2 U-2 X-2 C-1 V-1 W-1 D-0 E-0 F-0 G-0 H-0 I-0 J-0 K-0 L-0 M-0 N-0 O-0 P-0 Q-0 R-0 S-0 T-0 Y-0 Z-0

#include <stdio.h>
#include <ctype.h>
#include <string.h>

#define MAX_LEN 81

int main() {
    char str[MAX_LEN];
    int freq[26] = {0}; // 存储每个字母出现的次数，下标对应字母表顺序

    fgets(str, MAX_LEN, stdin);
    int len = strlen(str);

    for (int i = 0; i < len; i++) {
        if (isalpha(str[i])) { // 判断是否为字母
            freq[toupper(str[i]) - 'A']++; // 统计字母出现的次数
        }
    }

    // 冒泡排序，按字母出现的次数从高到低排序
    for (int i = 0; i < 25; i++) {
        for (int j = 0; j < 25 - i; j++) {
            if (freq[j] < freq[j+1] || (freq[j] == freq[j+1] && j > j+1)) {
                int temp = freq[j];
                freq[j] = freq[j+1];
                freq[j+1] = temp;
            }
        }
    }

    // 输出结果
    for (int i = 0; i < 26; i++) {
        if (freq[i] != 0) {
            printf("%c-%d ", 'A'+i, freq[i]);
        } else {
            printf("%c-%d ", 'A'+i, 0);
        }
    }
    return 0;
}