Add kattis prob ivana in python3 + notes

kattis/ivana/notes.md

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+If Ivana takes one integer, what is left is an integer chain from A to B. The next player will always need to take an integer from the end, leaving a chain of either A+1 to B, or A to B-1. We can do dynamic programming based on this.

kattis/ivana/solution.py

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+n = int(input())
+d = list(map(int, input().split()))
+
+dp = [[0 for _ in range(2*n)] for _ in range(2*n)]
+for v in range(n):
+    ans = d[v]%2
+    dp[v][v] = ans
+    dp[v+n][v+n] = ans
+
+for c in range(2, n+1):
+    for a in range(2*n):
+        b = a + c - 1
+        if b >= 2*n:
+            break
+        dp[a][b] = max(dp[a][a] - dp[a+1][b], dp[b][b] - dp[a][b-1])
+
+ans = 0
+for i in range(n):
+    if dp[i][i] > dp[i+1][i+n-1]:
+        ans += 1
+
+print(ans)