UVA-10719 Quotient Polynomial
2016-08-10 20:29
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2016-08-10
题目大意:多项式 p(x)=(x-k)q(x)+r,给出 k 和 p(x)求 q(x) 和 r。
解题思路:待定系数,推得 b[i] = a[i] - k * b[i-1],其中a 为 p 的系数,b 为 q 的系数。
k 是 (x-k) 的 k,不是 k 次方,样例害死人!!!
UVA - 10719 Quotient Polynomial
题目大意:多项式 p(x)=(x-k)q(x)+r,给出 k 和 p(x)求 q(x) 和 r。解题思路:待定系数,推得 b[i] = a[i] - k * b[i-1],其中a 为 p 的系数,b 为 q 的系数。
k 是 (x-k) 的 k,不是 k 次方,样例害死人!!!
#include <iostream> #include <cstdio> using namespace std; int a[100000],b[100000]; int main() { int k; int r; char c; while ( scanf("%d", &k) != EOF ) { k *= (-1); int n = 1; while ( scanf("%d%c", &a , &c) ) { if ( c == '\n' ) break; n++; } b[0] = 0; cout << "q(x):"; for (int i = 1; i < n; i++) { b[i] = a[i] - k*b[i-1]; cout << " " << b[i]; } r = a - k*b[n-1]; cout << endl << "r = " << r << endl << endl; } return 0; }
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