UVA 1546 - Complete the sequence!

题目链接

题意：给定多项式前s项，求出后c项，要求尽量小

思路：利用差分法，对原序列求s - 1次差分，就可以发现规律，然后对于每多一项，就逆推回去即可

代码：

#include 
  
   
#include 
   
     const int N = 205; int t, s, c, a[N][N]; int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &s, &c); memset(a, 0, sizeof(a)); for (int i = 0; i < s; i++) scanf("%d", &a[0][i]); for (int i = 1; i < s; i++) for (int j = 0; j < s - i; j++) a[i][j] = a[i - 1][j + 1] - a[i - 1][j]; for (int i = 0; i < c; i++) { for (int j = s - 1; j >= 0; j--) a[j][s - j + i] = a[j + 1][s - j - 1 + i] + a[j][s - j - 1 + i]; printf("%d%c", a[0][s + i], (i == c - 1 ? '\n' : ' ')); } } return 0; }