# Greedy + dp; c++; O(n(log(n)))

• one observation:
the pick at the end should always be n - (2^k - 1);
We can prove this with some tricky math....

``````class Solution {
public:
int getMoneyAmount(int n) {
int re[n+1];
int count[n+1];
re[0] = 0; count[0] = 0;
re[1] = 0;count[1] = 0;
re[2] = 1;count[2] = 1;
re[3] = 2;count[3] = 1;
re[4] = 4;count[4] = 2;

for(int i = 5; i <= n; i++)
{
int mintmp = INT_MAX;
int sum = 0;
for(int j = 2; j <= i; j*=2)
{

int tmp = max(i-j+1 + sum, re[i-j] + i-j+1);
sum+=i-j+1;
if(tmp < mintmp)
{mintmp = tmp;}

}
re[i] = mintmp;
}

return re[n];

}
};
``````

• @jersey Could you let us know the math behind the scene? :)

• @jersey Your solution fails for example n=124. Your output is 557, correct is 555.

• I don't think there is any trick for this problem, you have to enumerate all.

In DP, I tried to calculate cost for range [1+x, n+x] from [1, n] but failed; then I realized there is no trick, have to enumerate...

• @StefanPochmann Thanks, added this test case.

• @xing2 same here. I tried maintaining how many counts are used in [1..n] so [1+x,n+x] would be x*count + [1..n] with the same selection. But after failing test case 17 it took me hours to realize the selection of numbers may change...

• I did the same thing. But I want to blame the judger :) because my first version, which is finished within 2 minutes, correct but using top-down DP, failed due to TLE. Then I thought there must be relationship between [1+x, n+x] and [1, n] (and my instinct tell me that is wrong!) The intuition is simple: when you have larger numbers (with x), you want to use less times to guess, however it is super hard to provide a counter example.

• I did exactly the same thing yesterday however failed at n = 124. The correct is 555 and ours 557. And I don't know where the problem was yet.

• @xing2 Well depends on what you mean with "relationship". But here's a small example showing it's at least not trivial: The range [1,8] has a best worst case of \$12, namely \$5+\$7. And for the range [1001,1008], which is the same range but 1000 higher, it's \$3009. Namely \$1005+\$1003+\$1001. It isn't \$1005+\$1007 because that's one fewer guess, and up there, the number of guesses is most important.

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