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:
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]. |
I hereby claim:
To claim this, I am signing this object:
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]); | |
} |
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; |
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 |
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; |
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 |
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]; | |
7 7 | |
....... | |
..####. | |
.#...#. | |
.#.#.#. | |
.#...#. | |
.#####. | |
....... |
!A!B | |
AAA! | |
!AAA | |
C!A! |