diff --git a/PATAdvanced/1008.c b/PATAdvanced/1008.c
new file mode 100644
index 0000000..81b2376
--- /dev/null
+++ b/PATAdvanced/1008.c
@@ -0,0 +1,47 @@
+/**
+ * 1008. Elevator (20)
+ * 
+ * The highest building in our city has only one elevator. A request list is 
+ * made up with N positive numbers. The numbers denote at which floors the
+ * elevator will stop, in specified order. It costs 6 seconds to move the 
+ * elevator up one floor, and 4 seconds to move down one floor. The elevator 
+ * will stay for 5 seconds at each stop.
+ * 
+ * For a given request list, you are to compute the total time spent to fulfill 
+ * the requests on the list. The elevator is on the 0th floor at the beginning 
+ * and does not have to return to the ground floor when the requests are 
+ * fulfilled.
+ * 
+ * Input Specification:
+ * 
+ * Each input file contains one test case. Each case contains a positive integer
+ * N, followed by N positive numbers. All the numbers in the input are less than
+ * 100.
+ * 
+ * Output Specification:
+ * 
+ * For each test case, print the total time on a single line.
+ * 
+ * Sample Input:
+ * 3 2 3 1
+ * Sample Output:
+ * 41
+ */
+#include <stdio.h>
+int main()
+{
+    int N, time = 0, cur, pre = 0;
+    
+    scanf("%d", &N);
+    while(N--)
+    {
+        scanf("%d", &cur);
+        if(cur > pre)   time += (cur - pre) * 6 + 5;
+        else            time += (pre - cur) * 4 + 5;
+        pre = cur;
+    }
+    
+    printf("%d", time);
+    
+    return 0;
+}
diff --git a/PATAdvanced/1009.c b/PATAdvanced/1009.c
new file mode 100644
index 0000000..2b7bfc4
--- /dev/null
+++ b/PATAdvanced/1009.c
@@ -0,0 +1,50 @@
+/**
+ * 1009. Product of Polynomials (25)
+ * 
+ * This time, you are supposed to find A*B where A and B are two polynomials.
+ * 
+ * Input Specification:
+ * 
+ * Each input file contains one test case. Each case occupies 2 lines, and each 
+ * line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, 
+ * where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2,
+ * ..., K) are the exponents and coefficients, respectively. It is given that 1 
+ * <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.
+ * 
+ * Output Specification:
+ * 
+ * For each test case you should output the product of A and B in one line, with
+ * the same format as the input. Notice that there must be NO extra space at the
+ * end of each line. Please be accurate up to 1 decimal place.
+ * 
+ * Sample Input
+ * 2 1 2.4 0 3.2
+ * 2 2 1.5 1 0.5
+ * Sample Output
+ * 3 3 3.6 2 6.0 1 1.6
+ */
+
+#include <stdio.h>
+int main()
+{
+    int N, exp, count = 0;
+    float coef, A[1001] = {0}, B[1001] = {0}, MUL[2001] = {0};
+
+    scanf("%d", &N);
+    while(N--){ scanf("%d %f", &exp, &coef); A[exp] = coef; }
+    scanf("%d", &N);
+    while(N--){ scanf("%d %f", &exp, &coef); B[exp] = coef; }
+
+    for(int i = 0; i < 1001; i++)
+        for(int j = 0; j < 1001; j++)
+            MUL[i + j] += A[i] * B[j];
+    
+    for(int i = 0; i < 2001; i++)
+        if(MUL[i]) count++;
+    
+    printf("%d", count);
+    for(int i = 2000; i >= 0; i--) if(MUL[i]) 
+        printf(" %d %.1f", i, MUL[i]);
+    
+    return 0;
+}