[UVA][數學] 11170 - Cos(NA)
Problem F
Cos(NA)
Input: Standard Input
Output: Standard Output
Have you ever looked at formulae of the form Cos(NA)? If you haven’t looked at them yet, just look at them now:
These formulae will make you believe that any Cos(NA) can be expanded in an expression which contains only one function Cos(A) and all the coefficients are also integers. In this problem your job is to find such a formula for Cos(NA) given the value of N.
Input
The input file contains at most 50 lines of inputs. Each line contains a positive integer N (N<50). Input is terminated by a line containing a single zero.
Output
For each line of input except the last one you should produce one line of output. This line should contain the formula (As described in the problem statment) for Cos(NA). You don’t need to print any redundant things in the output such as (a) Printing operators in two consecutive places (b) Printing the exponent when it is 1 (c) Printing the coefficient when it is 1 (d) Just look at the output for sample input for details.
Sample Input Output for Sample Input
2340 |
2Cos^2(A)-1 4Cos^3(A)-3Cos(A) 8Cos^4(A)-8Cos^2(A)+1
|
Problem setter: Shahriar Manzoor
Moderator: Derek Kisman
討論區的說明
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
so cos(a+b)+cos(a-b) = 2 cos(a)cos(b).
Now let a = (N-1)A, b = A, and you get the formula
cos(NA) = 2cos((N-1)A) * cos(A) - cos((N-2)A) and it is clear that cos(NA) is a polynomial of degree N in cos(A)
int main() {
long long coef[50][50] = {};
coef[1][1] = 1;
coef[2][0] = -1, coef[2][2] = 2;
int i, j, k, n;
for(i = 3; i < 50; i++) {
for(j = 0; j <= i; j++) {
if(j)
coef[i][j] += 2*coef[i-1][j-1];
coef[i][j] -= coef[i-2][j];
}
}
while(scanf("%d", &n) == 1 && n) {
int f = 0;
for(i = n; i >= 0; i--) {
if(coef[n][i]) {
if(f && coef[n][i] > 0)
putchar('+');
f = 1;
if(coef[n][i] == 1 && i) {
}
else if(coef[n][i] == -1 && i)
printf("-");
else
printf("%lld", coef[n][i]);
if(i > 1)
printf("Cos^%d(A)", i);
else if(i == 1)
printf("Cos(A)");
}
}
puts("");
}
return 0;
}
/*
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
so cos(a+b)+cos(a-b) = 2 cos(a)cos(b).
Now let a = (N-1)A, b = A, and you get the formula
cos(NA) = 2cos((N-1)A) * cos(A) - cos((N-2)A) and it is clear that cos(NA) is a polynomial of degree N in cos(A)
*/