UVA 10105 Polynomial Coefficients(数论)
| Polynomial coefficients |
The Problem
The problem is to calculate the coefficients in expansion of polynomial (x1+x2+...+xk)n.
The Input
The input will consist of a set of pairs of lines. The first line of the pair consists of two integers n andk separated with space (0<K,N<13). This integers define the power of the polynomial and the amount of the variables. The second line in each pair consists of k non-negative integers n1, ..., nk, where n1+...+nk=n.
The Output
For each input pair of lines the output line should consist one integer, the coefficient by the monomialx1n1x2n2...xknk in expansion of the polynomial (x1+x2+...+xk)n.
Sample Input
2 2 1 1 2 12 1 0 0 0 0 0 0 0 0 0 1 0
Sample Output
2 2
题意：给定n,k。在给k个数nk。(x1 +x2 + x3 +...+xk) ^ n的展开式中x1^n1*x2^n2*...*xk*nk的系数。
思路：对于(x1 +x2 + x3 +...+xk) ^ n可以看成(Sk - 1 + xk) ^ n然后对于Sk - 1又能看成Sk - 2, xk -1。然后系数为Cn nk.这样一直求到n为0为止
代码：
#include <stdio.h>
#include <string.h>
const int N = 15;
int c[N][N], n, k, nk[N];
int cal(int n, int m) {
int a = 1, b = 1, num = m;
for (int i = 0; i < num; i ++) {
a *= n; n --;
b *= m; m --;
}
return a / b;
}
void init() {
for (int i = 1; i <= 12; i ++) {
for (int j = 0; j <= i; j ++) {
c[i][j] = cal(i, j);
}
}
}
int solve() {
int ans = 1;
for (int i = k; i >= 0; i --) {
ans *= c[n][nk[i]];
n -= nk[i];
if (n == 0)
return ans;
}
return ans;
}
int main() {
init();
while (~scanf("%d%d", &n, &k)) {
for (int i = 1; i <= k; i ++)
scanf("%d", &nk[i]);
printf("%d\n", solve());
}
return 0;
}