tchomphoochan

Require Import List Lia Arith.
Import ListNotations.

Fixpoint prefix_sum (l : list nat) :=
  match l with
  | nil => nil
  | cons d l' => cons d (map (fun x => x+d) (prefix_sum l'))
  end.

Goal prefix_sum [3; 2; 7; 6; 8] = [3; 5; 12; 18; 26].

tchomphoochan

Keybase proof

I hereby claim:

• I am tchomphoochan on github.
• I am tcpc (https://keybase.io/tcpc) on keybase.
• I have a public key whose fingerprint is FAD6 4375 6D8E E9D9 E4A8 595C 5A05 4AC1 4D33 3C67

To claim this, I am signing this object:

tchomphoochan

int A[MAX_N]; // assume data is in A[1..n]
int dp[MAX_N];

int ans = 0;
for (int i = 1; i <= n; ++i) {
    dp[i] = 1;
    for (int j = 1; j < i; ++j) {
        if (A[j] < A[i])
            dp[i] = max(dp[i], 1+dp[j]);
    }

tchomphoochan

int A[MAX_N];
int dp[MAX_N]; // dp[0] = 0

int ans = -1e9;
for (int i = 1; i <= n; ++i) {
    dp[i] = max(A[i], dp[i-1]+A[i]);
    ans = max(ans, dp[i]);
}

cout << ans << endl;

tchomphoochan

char A[MAX_N][MAX_M]; // assume the board's data is in A[1..n][1..m]
int dp[MAX_N][MAX_M]; // set everything to 0 (so dp[0][0..m] and dp[0..n][0] will be 0)

for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
        if (A[i][j] == '#')
            dp[i][j] = 0;
        else if (i == 1 && j == 1)
            dp[i][j] = 1;
        else

tchomphoochan

char A[MAX_N][MAX_M]; // assume the board's data is in A[1.n][1..m]
int dp[MAX_N][MAX_M]; // initially set all to -1 to mark as "not done yet"

int count_paths(int i, int j) {
    if (i < 1 || j < 1)
        return 0;
    if (A[i][j] == '#')
        return 0;
    if (i == 1 && j == 1)
        return 1;

tchomphoochan

int ans[MAX_N];
ans[0] = 0; // base case

for (int i = 1; i <= n; ++i) {
    ans[i] = 1e9; // don't know a way to solve yet
    if (i-1 >= 0) // try using 1-baht coin if possible
        ans[i] = min(ans[i], 1+ans[i-1]);
    if (i-3 >= 0) // try using 3-baht coin if possible
        ans[i] = min(ans[i], 1+ans[i-3]);
    if (i-4 >= 0) // try using 4-baht coin if possible

tchomphoochan

int memo[MAX_N]; // initially set all to -1 to mark as "not done yet"

int change(int i) {
    // base cases:
    if (i == 0) return 0;
    if (i < 0) return 1e9; // 1e9 = 1,000,000,000
    // already solved cases:
    if (memo[i] != -1)
        return memo[i];
    

tchomphoochan

7 7
.......
..####.
.#...#.
.#.#.#.
.#...#.
.#####.
.......

tchomphoochan

!A!B
AAA!
!AAA
C!A!