Code Chef DSA Learning Series : Multiple of 3 - c++

Question

Consider a very long K-digit number N with digits d0, d1, ..., dK-1 (in decimal notation; d0 is the most significant and dK-1 the least significant digit). This number is so large that we can't give it to you on the input explicitly; instead, you are only given its starting digits and a way to construct the remainder of the number.
Specifically, you are given d0 and d1; for each i ≥ 2, di is the sum of all preceding (more significant) digits, modulo 10 — more formally, the following formula must hold:
Determine if N is a multiple of 3.
Input
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first and only line of each test case contains three space-separated integers K, d0 and d1.
Output
For each test case, print a single line containing the string "YES" (without quotes) if the number N is a multiple of 3 or "NO" (without quotes) otherwise.
Constraints
1 ≤ T ≤ 1000
2 ≤ K ≤ 1012
1 ≤ d0 ≤ 9
0 ≤ d1 ≤ 9
Example
Input:
3
5 3 4
13 8 1
760399384224 5 1
Output:
NO
YES
YES
Explanation
Example case 1: The whole number N is 34748, which is not divisible by 3, so the answer is NO.
Example case 2: The whole number N is 8198624862486, which is divisible by 3, so the answer is YES.
Question Ended
In this question, in the example test case given, we have k=760399384224, d0=5, and d1=1. Now we know that a number is multiple of 3 if the sum of it’s digits is a multiple of 3. So applying it here, we separate the number n into 3 parts,
Part 1: First 3 digits -> 516 , (5+1+6) mod 3 ==0, so rem1 = 0.
Part 2: Next will be (k-3)/4 times repetition of (2486) ,or,rem2 = ((2+4+8+6)*((k-3)/4))%3= 1
Part 3: the last (k-3)%4 =1 digits which will be 2 (from 2486 repetition) , so rem3 = 2%3=2
So the final answer should be (rem1+rem2+rem3)%3
and I wrote the following code for this logic:
#include<iostream>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
ll k;
cin>>k;
int d0,d1;
cin>>d0>>d1;
int d2 = (d0+d1)%10;
int d[4];
d[0] = (d0+d1+d2)%10;
d[1] = (2*d[0])%10;
d[2] = (2*d[1])%10;
d[3] = (2*d[2])%10;
ll rem1 = (d0+d1+d2)%3,rem2,rem3=0;
rem2 = (20*(((k-3)/4)%3))%3;
ll x = (k-3)%4;
if(x!=0)
{
for(int i=0; i<x; ++i)
rem3+=d[i];
rem3 = rem3%3;
}
else
rem3 =0;
if(k==2)
{
rem1 = (d0+d1)%3;
rem2=0;
rem3=0;
}
if((rem1+rem2+rem3)%3==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
}
Now they’re giving me WA in their test cases. And I cant think of a possible test cases which doesn’t work with this code. So somebody help me out please.

Answer 1

Here's an apparent mismatch. You can probably use this JavaScript code to find more:
function f(k, d0, d1){
let d2 = (d0+d1)%10;
let d = new Array(4).fill(0);
d[0] = (d0+d1+d2)%10;
d[1] = (2*d[0])%10;
d[2] = (2*d[1])%10;
d[3] = (2*d[2])%10;
let rem1 = (d0+d1+d2)%3, rem2, rem3=0;
rem2 = (20*((~~((k-3)/4))%3))%3;
let x = (k-3)%4;
if(x!=0)
{
for(let i=0; i<x; ++i)
rem3+=d[i];
rem3 = rem3%3;
}
else
rem3 =0;
if(k==2)
{
rem1 = (d0+d1)%3;
rem2=0;
rem3=0;
}
if((rem1+rem2+rem3)%3==0)
return true
return false
}
function bruteForce(k, d0, d1){
let d = (d0 + d1) % 10;
let n = 10 * d0 + d1;
for (let i=3; i<=k; i++){
n = 10 * n + d;
d = (2 * d) % 10;
}
return [n % 3 == 0, n];
}
let str = "";
str += "Examples:\n";
str += "(5, 3, 4)\n" + bruteForce(5, 3, 4) + "\n\n";
str += "(13, 8, 1)\n" + bruteForce(13, 8, 1) + "\n\n"
var k = 7;
var mismatch = false;
for (let i=1; i<10; i++){
for (let j=0; j<10; j++){
const _f = f(k, i, j);
const _bruteForce = bruteForce(k, i, j);
if (_bruteForce[0] != _f){
str += "Mismatch:\n" +
`(${ k }, ${ i }, ${ j })\n` +
"f: " + _f +
"\nbruteForce: " + _bruteForce + "\n\n";
mismatch = true;
}
if (mismatch)
break;
}
if (mismatch)
break;
}
console.log(str);

Answer 2

#include <stdio.h>
void isMulti3(long long int K, long long int d0, long long int d1) {
long long int mod1 = (d0 + d1) % 10;
long long int sum_mod1 = d0 + d1 + mod1;
long long int rep = (K-3)/4;
long long int mod2[4];
mod2[0] = (2*mod1) % 10;
mod2[1] = (4*mod1) % 10;
mod2[2] = (8*mod1) % 10;
mod2[3] = (6*mod1) % 10;
long long int sum_mod2 = 0;
for(int i = 0; i < 4; i++) {
sum_mod2 += mod2[i];
}
long long int unrep = (K - 3) % 4;
long long int sum_mod3 = 0;
for(int i = 0; i < unrep; i++) sum_mod3 += mod2[i];
long long int isMod1 = sum_mod1 % 3;
long long int isMod2 = ((rep%3)*(sum_mod2%3)) % 3;
long long int isMod3 = sum_mod3 % 3;
if((isMod1 + (isMod2 + isMod3)%3)% 3 == 0) printf("YES\n");
else printf("NO\n");
}
int main() {
int T;
scanf("%d", &T);
for(int i = 0; i < T; i++) {
long long int K;
long long int d0, d1;
scanf("%lld %lld %lld", &K, &d0, &d1);
if(K == 2) {
if((d0 + d1) % 3 == 0) printf("YES\n");
else printf("NO\n");
}
else isMulti3(K, d0, d1);
}
}
I think you are the same guy who asked the question on the codechef discussion portal.
I answered it there , and I am sharing the link.
[https://discuss.codechef.com/t/dsa-learning-series-multiple-of-3/77174/4?u=nazishkaunain][1]
So the point where you are mistaken is:
You might use the fact:
(a+b+c)mod 3 != a mod 3 + b mod 3 + c mod 3;
But (a + b + c) mod 3 = (a mod 3 + ( b mod 3 + c mod 3) mod 3) mod 3;
And a number n ( n = a + b + c) is divisible by 3 only if n mod 3 = 0 => (a + b + c) mod 3 = 0.