Beginner C template is not very good at use! Solving!! Thank you!

Isn't this a function template for you? Take the following for example:

inline t1 max(t1 a,t1 b)if(a>b)

return a;

elsereturn b;

The call part is:

vectorivec( iarray, iarray+7 );

imax=max(max(ivec),max(iarray,7));

coutreturn a;

elsereturn b;

The others are of course similar, so I won't go into details! However, the keyword typename can also be replaced with class

Here's an example:

Spend less than 3 minutes: base = 2

Spend more than 3 minutes per minute: above = 1

The amount of money you have: total = 4

Obviously, it is more cost-effective to play for 3 minutes with 2 bucks from base than for 1 dollar per minute above, so using all the money for 3 minutes can give full play to the maximum value, there are a total of 4 bucks, and if you play two or three minutes, you can play for 6 minutes, so the output is 6.

The program can be written like this (greedy algorithm): first determine which is the best deal. If the base is cost-effective, all the money will be used to play the base, and the remaining money will not be enough to use the base on the above; If above is cost-effective, just hit the base once and then hit both above, and if the money left is not enough for the above but enough base, go to the base.

I really don't understand what you're going to do, so I'll be blind and touch the elephant once:

y=√(x⁴+ 3x² -6x + 10) －x⁴- 3x² +2x + 5)

y ' 4x³ +6x - 6) /x⁴+ 3x² -6x + 10) －4x³ -6x + 2) /x⁴- 3x² +2x + 5)

10 [1 - 3x/10 + 21x²/200 + 63x³/2000 + 863x^4/16000 +

x - 3x +2x + 5) - McLaughlin.

5 [1 + 1x/5 - 8x²/25 + 8x³/125 + 9x⁴/250 + 83x^5 /6250 + 213x^6/31250 +

x + 3x -6x + 10) x - 3x +2x + 5) - the denominator is rationalized.

x⁴+ 3x² -6x + 10) +x⁴- 3x² +2x + 5)] 6x² -8x + 5]

x + 3x -6x + 10) x -2x + 2) * (x + 2x + 5) - factorization.

x - 3x +2x + 5) cannot be decomposed in the range of real numbers.

If you don't know hi yet

You didn't give a specific value for the parameter, so I had to make it up.

function hh

fsolve(@myfun,[100,10])

function f=myfun(x)

t=x(1);t=x(2);

c1=1;c2=9;a=8;b=3;c=4;h=5;k=6;m=7;

f(1)=c2*(t-t)*(a+1/2*b*(t-t)+1/3*c*(t^2+t*t+t^2));

f(2)=(c*t^2+b*t+a)*(c1+c2+h/m+k/m^2)*exp(m*t)-(h+k*t)/m-k/m^2-c2*(t-t));

Result: ans =

Or so, simpler :

c1=1;c2=9;a=8;b=3;c=4;h=5;k=6;m=7;

f=@(x) [c2*(x(1)-x(2))*a+1/2*b*(x(1)-x(2))+1/3*c*(x(1)^2+x(1)*x(2)+x(2)^2));

c*x(2)^2+b*x(2)+a)*(c1+c2+h/m+k/m^2)*exp(m*x(2))-h+k*x(2))/m-k/m^2-c2*(x(1)-x(2)))

x=fsolve(f,[100,10])

t=x(1),t=x(2)

Result: x =

t =t =