zte's code,first commit
Change-Id: I9a04da59e459a9bc0d67f101f700d9d7dc8d681b
diff --git a/ap/lib/libssl/openssl-1.1.1o/crypto/bn/bn_kron.c b/ap/lib/libssl/openssl-1.1.1o/crypto/bn/bn_kron.c
new file mode 100644
index 0000000..c1e09d2
--- /dev/null
+++ b/ap/lib/libssl/openssl-1.1.1o/crypto/bn/bn_kron.c
@@ -0,0 +1,140 @@
+/*
+ * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
+ *
+ * Licensed under the OpenSSL license (the "License"). You may not use
+ * this file except in compliance with the License. You can obtain a copy
+ * in the file LICENSE in the source distribution or at
+ * https://www.openssl.org/source/license.html
+ */
+
+#include "internal/cryptlib.h"
+#include "bn_local.h"
+
+/* least significant word */
+#define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
+
+/* Returns -2 for errors because both -1 and 0 are valid results. */
+int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
+{
+ int i;
+ int ret = -2; /* avoid 'uninitialized' warning */
+ int err = 0;
+ BIGNUM *A, *B, *tmp;
+ /*-
+ * In 'tab', only odd-indexed entries are relevant:
+ * For any odd BIGNUM n,
+ * tab[BN_lsw(n) & 7]
+ * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
+ * Note that the sign of n does not matter.
+ */
+ static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
+
+ bn_check_top(a);
+ bn_check_top(b);
+
+ BN_CTX_start(ctx);
+ A = BN_CTX_get(ctx);
+ B = BN_CTX_get(ctx);
+ if (B == NULL)
+ goto end;
+
+ err = !BN_copy(A, a);
+ if (err)
+ goto end;
+ err = !BN_copy(B, b);
+ if (err)
+ goto end;
+
+ /*
+ * Kronecker symbol, implemented according to Henri Cohen,
+ * "A Course in Computational Algebraic Number Theory"
+ * (algorithm 1.4.10).
+ */
+
+ /* Cohen's step 1: */
+
+ if (BN_is_zero(B)) {
+ ret = BN_abs_is_word(A, 1);
+ goto end;
+ }
+
+ /* Cohen's step 2: */
+
+ if (!BN_is_odd(A) && !BN_is_odd(B)) {
+ ret = 0;
+ goto end;
+ }
+
+ /* now B is non-zero */
+ i = 0;
+ while (!BN_is_bit_set(B, i))
+ i++;
+ err = !BN_rshift(B, B, i);
+ if (err)
+ goto end;
+ if (i & 1) {
+ /* i is odd */
+ /* (thus B was even, thus A must be odd!) */
+
+ /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
+ ret = tab[BN_lsw(A) & 7];
+ } else {
+ /* i is even */
+ ret = 1;
+ }
+
+ if (B->neg) {
+ B->neg = 0;
+ if (A->neg)
+ ret = -ret;
+ }
+
+ /*
+ * now B is positive and odd, so what remains to be done is to compute
+ * the Jacobi symbol (A/B) and multiply it by 'ret'
+ */
+
+ while (1) {
+ /* Cohen's step 3: */
+
+ /* B is positive and odd */
+
+ if (BN_is_zero(A)) {
+ ret = BN_is_one(B) ? ret : 0;
+ goto end;
+ }
+
+ /* now A is non-zero */
+ i = 0;
+ while (!BN_is_bit_set(A, i))
+ i++;
+ err = !BN_rshift(A, A, i);
+ if (err)
+ goto end;
+ if (i & 1) {
+ /* i is odd */
+ /* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
+ ret = ret * tab[BN_lsw(B) & 7];
+ }
+
+ /* Cohen's step 4: */
+ /* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
+ if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
+ ret = -ret;
+
+ /* (A, B) := (B mod |A|, |A|) */
+ err = !BN_nnmod(B, B, A, ctx);
+ if (err)
+ goto end;
+ tmp = A;
+ A = B;
+ B = tmp;
+ tmp->neg = 0;
+ }
+ end:
+ BN_CTX_end(ctx);
+ if (err)
+ return -2;
+ else
+ return ret;
+}