HDU 4927 Big Data

The question is simple:

For the number of n characters in length, perform n-1 times to generate a new series:

B1 = a2-a1 b2= a3-a2 B3 = a4-a3

C1 = b2-b1 C2 = b3-b2

Ans = c2-c1

Finally, the formula is as follows: Yang Hui triangle of N's row

For example:

3

1 2 3

Ans = 1*1-2*2 + 1*3 = 0

4

1 5 7 2

Ans =-1*1 + 3*5-3*7 + 1*2 =-5

You can optimize each number in the Yang Hui triangle. The last number can be multiplied and divided by the previous one.

Java uses the number of seconds... But not Java --#

Use C ++ to simulate large numbers ..

# Include "stdio. H "# include" string. H "int A [3010] ;__ int64 mark [3010], ANS [3010], C [3010]; int max (int A, int B) {if (a <B) return B; else return a;} void make_mul (int x) {int I; for (I = 1; I <= mark [0]; I ++) MARK [I] * = x; for (I = 1; I <= mark [0]; I ++) {mark [I + 1] + = mark [I]/1000000; Mark [I] % = 1000000;} while (MARK [MARK [0] + 1]! = 0) {mark [0] ++; Mark [MARK [0] + 1] = mark [MARK [0]/1000000; mark [MARK [0] % = 1000000 ;}} void make_div (int x) {int I; for (I = mark [0]; I> = 2; I --) {mark [I-1] + = (MARK [I] % x) x 1000000; Mark [I]/= x;} mark [1]/= X; while (MARK [MARK [0] = 0) MARK [0] --;} void make_add () {int I, OP; If (ANS [0]> 0) {for (I = 1; I <= mark [0]; I ++) {ans [I] + = mark [I]; ans [I + 1] + = ans [I]/1000000; ans [I] % = 1000000;} while (ANS [ans [0] + 1]! = 0) {ans [0] ++; ans [ans [0] + 1] = ans [ans [0]/1000000; ans [ans [0] % = 1000000;} return;} else {If (-ans [0]> MARK [0]) OP =-1; else if (-ans [0] <mark [0]) OP = 1; else {for (I = mark [0]; I> = 1; I --) if (MARK [I]> ans [I]) {op = 1; break;} else if (MARK [I] <ans [I]) {op =-1; break;} if (I = 0) {memset (ANS, 0, sizeof (ANS); ans [0] = 1; return ;}} if (OP = 1) {for (I = 1; I <= mark [0]; I ++) {ans [I] = mark [I]-ans [I]; If (ANS [I] <0) {Mark [I + 1] --; ans [I] + = 1000000 ;}} ans [0] =-ans [0]; while (ANS [ans [0] = 0) ans [0] --;} else {for (I = 1; I <=-ans [0]; I ++) {ans [I] = ans [I]-MARK [I]; If (ANS [I] <0) {ans [I + 1] --; ans [I] + = 1000000;} while (ANS [-ans [0] = 0) ans [0] ++ ;}} void make_red () {int I, OP; If (ANS [0] <0) {for (I = 1; I <= mark [0]; I ++) {ans [I] + = mark [I]; ans [I + 1] + = ans [I]/1000000; ans [I] % = 1000000 ;} while (ANS [-ans [0] + 1]! = 0) {ans [0] --; ans [-ans [0] + 1] = ans [-ans [0]/1000000; ans [-ans [0] % = 1000000;} return;} else {If (ANS [0]> MARK [0]) OP = 1; else if (ANS [0] <mark [0]) OP =-1; else {for (I = mark [0]; I> = 1; I --) if (MARK [I]> ans [I]) {op =-1; break;} else if (MARK [I] <ans [I]) {op = 1; break;} if (I = 0) {memset (ANS, 0, sizeof (ANS); ans [0] = 1; return ;}} if (OP = 1) {for (I = 1; I <= mark [0]; I ++) {ans [I] = ans [I]-MARK [I]; If (ANS [I] <0) {ans [I + 1] --; ans [I] + = 1000000;} while (ANS [ans [0] = 0) ans [0] --;} else {for (I = 1; I <= mark [0]; I ++) {ans [I] = mark [I]-ans [I]; If (ANS [I] <0) {mark [I + 1] --; ans [I] + = 1000000 ;}} ans [0] = mark [0]; while (ANS [ans [0] = 0) ans [0] --; ans [0] =-ans [0] ;}} int main () {int t, n, I, X, Y, J, m, OP; scanf ("% d", & T); While (t --) {scanf ("% d", & N); for (I = 1; I <= N; I ++) scanf ("% d ", & A [I]); memset (ANS, 0, sizeof (ANS); ans [0] = 1; if (I = 1) {printf ("% d \ n", a [1]); continue;} memset (mark, 0, sizeof (Mark); Mark [0] = 1; mark [1] = 1; if (N % 2 = 1) ans [1] = A [1]; else ans [1] =-A [1]; X = 1; y = n-1; for (I = 2; I <= N; I ++) {make_mul (y); y --; make_div (X ); X ++; memcpy (C, Mark, sizeof (Mark); make_mul (A [I]); If (N % 2 = 1) {if (I % 2 = 1) make_add (); else make_red ();} else {if (I % 2 = 0) make_add (); else make_red ();} memcpy (mark, C, sizeof (c);} If (ANS [0]> 0) {printf ("% i64d ", ans [ans [0]); for (I = ans [0]-1; I> = 1; I --) printf ("% 06i64d ", ans [I]); printf ("\ n");} else {printf ("-"); printf ("% i64d ", ans [-ans [0]); for (I =-ans [0]-1; I> = 1; I --) printf ("% 06i64d ", ans [I]); printf ("\ n") ;}} return 0 ;}