Hdu 4927 Series 1 (large number template addition, subtraction, multiplication, division)

Hdu 4927 Series 1 (large number template addition, subtraction, multiplication, division)

Series 1 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission (s): 1067 Accepted Submission (s): 390


Problem DescriptionLet A be an integral series {A1, A2,..., }.

The zero-order series of A is A itself.

The first-order series of A is {B1, B2,..., Bn-1}, where Bi = Ai + 1-Ai.

The ith-order series of A is the first-order series of its (I-1) th-order series (2 <= I <= n-1 ).

Obviusly, the (n-1) th-order series of A is a single integer. Given A, figure out that integer.
InputThe input consists of several test cases. The first line of input gives the number of test cases T (T <= 10 ).

For each test case:
The first line contains a single integer n (1 <= n <= 3000), which denotes the length of series.
The second line consists of n integers, describing A1, A2,..., An. (0 <= Ai <= 105)

OutputFor each test case, output the required integer in a line.
Sample Input

231 2 341 5 7 2

Sample Output

0-5

AuthorBUPT
Source2014 Multi-University Training Contest 6

The formula is as follows: C (n, n-1) = C (n, n) * n/1; C (n, N-2) = C (n, n-1) * (n-1)/2;

# Include "stdio. h "# include" string. h "# include" iostream "# include" algorithm "using namespace std; # define LL _ int64 # define N 300 # define M 100000000000 # define max (a, B) (a> B? A: B) int a [3010]; LL c [N], ans [N], cc [N]; // array size. The length of the first element of the array contains a large number, if the length is negative, it indicates that the large number is smaller than zero void mul (LL * c, int x) // a large number is multiplied by an integer {int I; for (I = 1; I <= c [0]; I ++) c [I] = c [I] * x; for (I = 1; I <= c [0]; I ++) {c [I + 1] + = c [I]/M; c [I] = c [I] % M ;} while (c [c [0] + 1]) {c [0] ++; c [c [0] + 1] = c [c [0]/M; c [c [0] % = M ;}} void div (LL * c, int y) // divide a large number by an integer {int I; for (I = c [0]; I> 1; I --) {c [I-1] + = (c [I] % y) * M; c [I]/= y;} c [1]/= y; while (c [c [0] = 0) c [0] --;} void add (LL * ans, LL * c) // add a large number to a large number, and save the result to ans {int f =-1, I; if (ans [0] <0) {if (-ans [0]> c [0]) f = 1; else if (-ans [0]
 
  
0; I --) if (ans [I]> c [I]) {f = 1; break;} else if (ans [I]
  
   
C [0]) f = 1; else if (ans [0]
   
    
0; I --) if (ans [I]> c [I]) {f = 1; break;} else if (ans [I]
    
     
0) {mul (c, n-I); div (c, I);} memcpy (cc, c, sizeof (c); mul (c, a [I]); if (flag = 1) add (ans, c); else sub (ans, c); flag * =-1; memcpy (c, cc, sizeof (cc);} if (ans [0] <0) {printf ("-"); ans [0] =-ans [0];} printf ("% I64d", ans [ans [0]); for (I = ans [0]-1; I> 0; I --) printf ("% 011I64d", ans [I]); printf ("\ n");} return 0 ;}