HackerEarth Recursive Sums problem solution

YASH PAL, 31 July 202413 February 2026

In this HackerEarth Recursive Sums problem solution Little Bear has received a home assignment to find the sum of all digits in a number N. Following his affinity towards single-digit numbers, he intends to repeatedly compute the sum of all digits until the sum itself becomes a single-digit number.

Can you write a program to compute the final single-digit sum?

As the number N is very big, it is given in the following run-length encoded format – N is represented as a sequence of M blocks, where each block i (0 ≤ i < M) is represented by two integers – (len[i], d[i]). This implies that the digit d[i] occurs a len[i] number of times.

HackerEarth Recursive Sums problem solution

HackerEarth Recursive Sums problem solution.

#include <bits/stdc++.h>
using namespace std;

#define MOD                     1000000007
#define pb(x)                   push_back(x)
#define mp(x,y)                 make_pair(x,y)
#define FF                      first
#define SS                      second
#define s(n)                    scanf("%d",&n)
#define sl(n)                   scanf("%lld",&n)
#define sf(n)                   scanf("%lf",&n)
#define ss(n)                   scanf("%s",n)
#define sc(n)                   {char temp[4]; ss(temp); n=temp[0];}
#define INF                     (int)1e9
#define LINF                    (long long)1e18
#define EPS                     1e-9
#define maX(a,b)                ((a)>(b)?(a):(b))
#define miN(a,b)                ((a)<(b)?(a):(b))
#define abS(x)                  ((x)<0?-(x):(x))

typedef long long ll;
typedef unsigned long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<int,PII> TRI;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<ll> vl;
typedef vector<PII> VII;
typedef vector<TRI> VT;

int n;
int TEST_NO;
void precompute() {
  
}
void read() {
  s(n);
  ll ans = 0;
  bool g0 = false;
  for (int i = 0; i < n; ++i) {
    ll len, d;
    cin >> len >> d;
    if(d > 0) g0 = true;
    ans += (len * d) % 9;
    if(ans >= 9) ans -= 9;
  }
  if(ans == 0 and g0) ans = 9;
  cout << ans << endl;
}
void preprocess() {
  
}
void solve() {
  
}
int main() {
  precompute();
  int t;
  s(t);
  for(TEST_NO = 1; TEST_NO <= t; TEST_NO ++) {
    read();
    preprocess();
    solve();
  }
  return 0;
}

Second solution

#include <iostream>
using namespace std;

#define ull unsigned long long

int cal(ull d) {
  if (d < 10) return d;
  int res = 0;
  while (d > 0) {
    res += d % 10;
    d /= 10;
  }
  return cal(res);
}

int main() {
    int cases, n, d;
    for (scanf("%d", &cases); cases--; ) {
      scanf("%d", &n);
      ull sum = 0, len;
      for (int i = 0; i < n; i++) {
        cin >> len >> d;
        sum += d * len;
        sum = cal(sum);
      }
      sum = cal(sum);
      cout << sum << endl;
    }
    return 0;
}

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