mccraveiro

Keybase proof

I hereby claim:

• I am mccraveiro on github.
• I am mccraveiro (https://keybase.io/mccraveiro) on keybase.
• I have a public key ASBQbnq4vEXhMwhqXuh-1Y48KwtEyHZ9zRID757etcWhxAo

To claim this, I am signing this object:

mccraveiro

const paymentMethods = [{
  supportedMethods: 'basic-card',
  data: {
    supportedNetworks: ['visa', 'mastercard', 'amex'],
  },
}]

// Checkout details
const details = {
  displayItems: [

mccraveiro

<Multi_key> <e> <A>     : "Á"   Aacute # LATIN CAPITAL LETTER A WITH ACUTE
<Multi_key> <e> <E>     : "É"   Eacute # LATIN CAPITAL LETTER E WITH ACUTE
<Multi_key> <e> <I>     : "Í"   Iacute # LATIN CAPITAL LETTER I WITH ACUTE
<Multi_key> <e> <O>     : "Ó"   Oacute # LATIN CAPITAL LETTER O WITH ACUTE
<Multi_key> <e> <U>     : "Ú"   Uacute # LATIN CAPITAL LETTER U WITH ACUTE

<Multi_key> <e> <a>     : "á"   aacute # LATIN SMALL LETTER A WITH ACUTE
<Multi_key> <e> <e>     : "é"   eacute # LATIN SMALL LETTER E WITH ACUTE
<Multi_key> <e> <i>     : "í"   iacute # LATIN SMALL LETTER I WITH ACUTE
<Multi_key> <e> <o>     : "ó"   oacute # LATIN SMALL LETTER O WITH ACUTE

mccraveiro

resources:
  Resources:
    IamRoleLambda:
      Type: 'AWS::IAM::Role'
      Properties:
        AssumeRolePolicyDocument:
          Version: '2012-10-17'
          Statement:
            -
              Effect: Allow

mccraveiro

// gist: https://gist.github.com/mccraveiro/45608fa73eb773c9f6b9
{
  "bitwise": true,
  "curly": true,
  "immed": true,
  "latedef": true,
  "nonbsp": true,
  "quotmark": "single",
  "undef": true,
  "unused": true,

mccraveiro

## Setup java

if [ `uname -m` == 'x86_64' ]; then
 PATH="/usr/lib64/jvm/java-7-oracle/jre/bin/"
 JAVA_HOME="/usr/lib64/jvm/java-7-oracle/"
else
 PATH="/usr/lib/jvm/java-7-oracle/jre/bin/"
 JAVA_HOME="/usr/lib/jvm/java-7-oracle/"
fi

mccraveiro

#include <stdio.h>
#include <math.h>

#define MAX 1000

typedef struct {
	int x, y;
} Point;

typedef struct {

mccraveiro

#define s(v) scanf("%d",&v);
#define a p[r][c]
int main(){int n,i,r,c,l,m,p[100][100];s(n)for(i=0;i<n;i++){s(l)m=0;for(r=0;r<l;r++)for(c=0;c<=r;c++){s(a)if(r!=0){a+=(c==0||p[r-1][c-1]<=p[r-1][c])?p[r-1][c]:p[r-1][c-1];if(r+1==l)m=a>m?a:m;}}printf("%d\n",m);}}

mccraveiro

#include <stdio.h>

int pd[13][99];

int permut(n, k) {

    if (pd[n][k] != -1)
        return pd[n][k];

    int i;