P12_RecursiveSum.py

# Little Bear has received a home assignment to find the sum of all digits in a number N. Following his affinity towards
# single digit number, 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 len[i] number of times.
#
# For example, {(2,1), (1,2), (2,9)} represents the number 11299.
#
# Input:
#
# The first line contains a single integer T, denoting the number of test cases.
# Each test case starts with a single integer M, which is the number of following run-length encoded blocks.
# Each of the next M lines contains two space separated integers: len[i] and d[i].
#
# Output:
#
# Print the single-digit sum for each of the T test cases in a new line.
#
# Constraints:
#
# 0 ≤ d[i] ≤ 9
# 1 ≤ len[i] ≤ 1015
# 1 ≤ M ≤ 104
# 1 ≤ T ≤ 10
#
# SAMPLE INPUT
# 3
# 3
# 2 1
# 1 2
# 2 9
# 1
# 8 1
# 3
# 2 5
# 1 4
# 1 5
#
# SAMPLE OUTPUT
# 4
# 8
# 1

testCases = int(input())
while testCases:
    testCases -= 1
    M = int(input())
    length = []
    d = []
    result = 0
    check = 0
    for _ in range(M):
        M -= 1
        check = [int(i) for i in input().split()]
        result += check[0] * check[1]
    if result < 9:
        print(result)
    else:
        while result > 9:
            result = list(str(result))
            check = [int(i) for i in result]
            result = sum(check)
        print(sum(check))