uva 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

2

3

4

0

2Cos^2(A)-1

4Cos^3(A)-3Cos(A)

8Cos^4(A)-8Cos^2(A)+1

 

---

Problem setter: Shahriar Manzoor

Moderator: Derek Kisman

输出cos函数的展开式，找到递推式，然后注意些细节。

暑假集训就这么开始了啊！

#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int INF = 1000000000;const int maxn = 50 + 5;LL ans[maxn][maxn];void pre(){    memset(ans, 0, sizeof(ans));    for(int i = 0;i < maxn;i++){        if(i%4 == 0) ans[0][i] = 1;        else if(i%4 == 2) ans[0][i] = -1;        else ans[0][i] = 0;    }    ans[1][1] = 1;    for(int i = 1;i < maxn;i++){        for(int j = 2;j < maxn;j++){            ans[i][j] = 2*ans[i-1][j-1] - ans[i][j-2];        }    }}int main(){    pre();    int n;    while(scanf("%d", &n)){        if(n == 0) break;        int tag = 1;        for(int i = n;i > 0;i--){            if(ans[i][n] != 0){                if(tag == 1){                    tag = 0;                }                else{                    if(ans[i][n] > 0)                        printf("+");                    else                        printf("-");                }                if(ans[i][n] != 1)                    printf("%lld", abs(ans[i][n]));                printf("Cos");                if(i != 1)                    printf("^%d", i);                printf("(A)");            }        }        if(ans[0][n] > 0)            printf("+%lld", ans[0][n]);        else if(ans[0][n] < 0)            printf("%lld", ans[0][n]);        printf("\n");    }    return 0;}