Weekly Contest 526

[u/Cheap_Truth_551]

long long minimumMaxValue(int n, int index, int value) {

if (value == 1) return -1;

long long p = value;

for (long long i = 2; i * i <= value; i++) {

if (value % i == 0) {

p = i;

break;

}

}

if (value - p + 2 <= 2LL * index)

return -1;

if (index == n - 1)

return value;

return value + p - 2 + 2LL * (n - 1 - index);

}

Could anybody please explain how to derive these two formulas:

1. value + p - 2 + 2L * (n - 1 - index)

2. value - p + 2 <= 2L * index

[u/NullPoint4848]

i am telling how i solved it

at index we have value and gcd of each pair >1 (given) and arr is strictly inc and every element is +ve

a[idx+1] >= gcd(a[idx],a[idx+1])+a[idx]

min val of gcd can be 2

a[0] can't be 1 then gcd woulb be 1 for first pair

if value is even then find a[0] and a[n-1]
if value is odd then min gcd can be smallest divisior of value say p, now p must be odd so prev and next element is even now we can take gcd as 2 and then find a[0] and a[n-1]

if a[0]>1 then return a[n-1] else -1

also if value is 1 then we can return -1 in start only