0
For unit place digit i.e 1 to 9 = 1 no
for tens place digit 10 to 19= 10+1( units place 1 for 11 number is added)
for 20 to 99 = 8 ones
total 20 no of ones upto two digit number.
these two digit no will repeat for all 100 to 499 i.e 100 to 199 20 ones, 200 to 299 20 ones, 300 to 399 20 ones and 400 to 499 20 ones
i.e 20*4 =80
including these from 100 to 199 the hundreth place one come 100 times
total = 20+80+100=200
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What is the program to figure out how many numbers from 0 to 9 are needed for 1 to 89 for example?
#include<iostream> #include<array> #include<sstream> std::array<int,10> get_frequency (int range_min, int range_max) { if (range_max<range_min) std::swap (range_min, range_max); std::array<int,10> digit {}; for (int count {range_min}; count<=range_max; ++count) { std::stringstream ss {}; ss << count; std::string s {}; ss >> s; for (auto c : s) { ++digit[c-'0']; } } return digit; } int main () { std::array<int,10> digit {}; digit = get_frequency(1, 89); std::cout << "In the range 1 to 89...\n"; for (int d {0}; d<10; ++d) { std::cout << "\tthe digit " << d << " appears " << digit[d] << " times.\n"; } } Output: In the range 1 to 89... the digit 0 appears 8 times. the digit 1 appears 19 times. the digit 2 appears 19 times. the digit 3 appears 19 times. the digit 4 appears 19 times. the digit 5 appears 19 times. the digit 6 appears 19 times. the digit 7 appears 19 times. the digit 8 appears 19 times. the digit 9 appears 9 times.
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