Data update

Task/Chinese-remainder-theorem/EasyLang/chinese-remainder-theorem.easy

@@ -1,35 +1,35 @@
-proc mul_inv a b . x1 .
-  b0 = b
-  x1 = 1
-  if b <> 1
-    while a > 1
-      q = a div b
-      t = b
-      b = a mod b
-      a = t
-      t = x0
-      x0 = x1 - q * x0
-      x1 = t
-    .
-    if x1 < 0
-      x1 += b0
-    .
-  .
+func mul_inv a b .
+   b0 = b
+   x1 = 1
+   if b <> 1
+      while a > 1
+         q = a div b
+         t = b
+         b = a mod b
+         a = t
+         t = x0
+         x0 = x1 - q * x0
+         x1 = t
+      .
+      if x1 < 0
+         x1 += b0
+      .
+   .
+   return x1
 .
 proc remainder . n[] a[] r .
-  prod = 1
-  sum = 0
-  for i = 1 to len n[]
-    prod *= n[i]
-  .
-  for i = 1 to len n[]
-    p = prod / n[i]
-    call mul_inv p n[i] h
-    sum += a[i] * h * p
-    r = sum mod prod
-  .
+   prod = 1
+   sum = 0
+   for i = 1 to len n[]
+      prod *= n[i]
+   .
+   for i = 1 to len n[]
+      p = prod / n[i]
+      sum += a[i] * (mul_inv p n[i]) * p
+      r = sum mod prod
+   .
 .
 n[] = [ 3 5 7 ]
 a[] = [ 2 3 2 ]
-call remainder n[] a[] h
+remainder n[] a[] h
 print h