24h購物| | PChome| 登入
2012-05-19 17:32:59| 人氣1,864| 回應0 | 上一篇 | 下一篇

[UVA] 498 - Polly the Polynomial

推薦 0 收藏 0 轉貼0 訂閱站台


 Polly the Polynomial 

Algebra! Remember algebra? There is a theory that as engineers progresses further and further in their studies, they lose basic math skills. This problem is designed to help you remember those basic algebra skills, make the world a better place, etc., etc.

Input

Your program should accept an even number of lines of text. Each pair of lines will represent one problem. The first line will contain a list of integers { tex2html_wrap_inline27 } which represent a set of coefficients to a polynomial expression. The order of the polynomial is n. The coefficients should be paired with the terms of the polynomial in the following manner:

displaymath31

The second line of text represents a sequence of values for x, { tex2html_wrap_inline35 }.

Output

For each pair of lines, your program should evaluate the polynomial for all the values of x ( tex2html_wrap_inline39 through tex2html_wrap_inline41 ) and output the resulting values on a single line.

Sample Input

-2
5 0 1 6
1 -1
7 6 -1

Sample Output

-2 -2 -2 -2
6 5 -2



#include <stdio.h>
int parseLine(char *str, int a[]) {
    int i, neg = 1, g = 0, tmp = 0, idx = 0;
    for(i = 0; str[i]; i++) {
        if(str[i] >= '0' && str[i] <= '9')
            tmp = tmp*10 + str[i]-'0', g = 1;
        else {
            if(str[i] == '-') {
                neg = -1;
            } else if(str[i] == '+') {
                neg = 1;
            } else if(g) {
                a[idx++] = tmp*neg;
                g = 0, tmp = 0, neg = 1;
            }
        }
    }
    if(g)
        a[idx++] = tmp*neg;
    return idx;
}
int pow(int x, int y) {
    if(y == 0)
        return 1;
    if(y&1)
        return pow(x*x, y>>1)*x;
    else
        return pow(x*x, y>>1);
}
int main() {
    char line[500];
    int c[500], x[500];
    while(gets(line)) {
        int ct, xt;
        ct = parseLine(line, c);
        gets(line);
        xt = parseLine(line, x);
        int sum, i, j;
        for(i = 0; i < xt; i++) {
            sum = 0;
            for(j = 0; j < ct; j++)
                sum += pow(x[i], ct-j-1)*c[j];
            if(i)
                putchar(' ');
            printf("%d", sum);
        }
        puts("");
    }
    return 0;
}

台長: Morris
人氣(1,864) | 回應(0)| 推薦 (0)| 收藏 (0)| 轉寄
全站分類: 不分類 | 個人分類: UVA |
此分類下一篇:[UVA][BIT] 11525 - Permutation
此分類上一篇:[UVA][priority_queue] 11926 - Multitasking

是 (若未登入"個人新聞台帳號"則看不到回覆唷!)
* 請輸入識別碼:
請輸入圖片中算式的結果(可能為0) 
(有*為必填)
TOP
詳全文