In this question pos vector is initially given value -1; the how can we change value later,also tell what exactly what method are we using in this question

#1

Ques. Divisible Subset
To find a non-empty subset of the given multiset with the sum of
elements divisible by the size of original multiset.
Hint:
a % N = x
b % N = x
Then, (b-a) % N = (b % N - a % N) = (x - x) = 0
Let’s denote a1 + a2 + … + ak by bk. So, we obtain:
b0 = 0
b1 = a1
b2 = a1 + a2
.
.
.
.
bN = a1 + a2 + a3 + a4 +…+ aN
so, aL + aL+1 + … + aR = bR - bL-1
Therefore, if there are two values with equal residues modulo N among
b0, b1, …, bN then we take the first one for L-1 and the second one for
R and the required subsegment is found.
There are N+1 values of bi and N possible residues for N. So,
according to the pigeonhole principle the required subsegment will
always exist.
Code
#include<bits/stdc++.h>
using namespace std;
#define N (1e6+10)
vector pos(N, -1);
vector v;
vector pre;
int main(){
int t,n,x,l = 0,r = 0;
scanf("%d",&t);
while(t–){
scanf("%d",&n);
for(int i = 0;i<n;i++){
scanf("%d",&x);
v.push_back(x % n);
}
//Pre vector denotes the prefix sum bi
pre.push_back(0);
for(int i = 0;i<n;i++)
pre.push_back((pre[i] + v[i]) % n);
//Position vector denotes the index of
//each segment sum
for(int i = 0;i<n+1;i++){
//If this segement sum is encountered first t
ime
if(pos[pre[i]] == -1){
pos[pre[i]] = i;
}
//If this segment sum has already been encoun
tered
//we have our answer.
//L will be its previous index and R current
index
else{
//pos[pre[i]] will give L-1
l = pos[pre[i]] + 1;
r = i;
break;
}
}
//Number of elements in the segment
//is (R-L+1)
printf("%d\n",r - l + 1);
//Print the index between l and r
for(int i = l;i<=r;i++)
printf("%d “,i);printf(”\n");
v.clear();
pre.clear();
for(int i = 0;i<=n;i++)pos[i] = -1;
}
return 0;
}

#2

@goyal431 vectors work just like arrays. If you want to rewrite any value later you can simply do it like pos[i] = 6; We are using the pigeonhole principle in this question, you can watch this https://www.youtube.com/watch?v=p5Cm_r4T1Rw&t=625s video for better understanding of this algorithm.

#3

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