1049. Counting Ones (30)

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (<=2³⁰).

Output Specification:

For each test case, print the number of 1's in one line.

Sample Input:

12

Sample Output:

5

来源： <http://www.patest.cn/contests/pat-a-practise/1049>

 

1. #include <string>
2. #include <iostream>
3. #include <math.h>
5. using namespace std;
6. string str;
7. int GetLeftNum(int a) {
8. int left = 0;
9. for (int i = 0; i < a; i++) {
10. left *= 10;
11. left += str[i] - '0';
12. }
13. return left;
14. }
15. int GetRightNum(int b) {
16. int right = 0;
17. for (int i = b+1; i < str.length(); i++) {
18. right *= 10;
19. right += str[i] - '0';
20. }
21. return right;
22. }
24. int main(void) {
26. cin >> str;
28. if (str.length() == 0) {
29. cout << 1;
30. return 0;
31. }
33. int oneCnt = 0;
34. int left = 0, right = 0;
35. for (int i = 0; i < str.length(); i++) {
36. left = GetLeftNum(i);
37. right = GetRightNum(i);
38. if (str[i] - '0' > 1) {
39. oneCnt += (left + 1)*pow(10, str.length() - i - 1);
40. }
41. else if (str[i] == '1') {
42. oneCnt+= (left)*pow(10, str.length() - i - 1)+right+1;
43. }
44. else {
45. oneCnt += (left)*pow(10, str.length() - i - 1);
46. }
47. }
48. cout << oneCnt;
49. return 0;
50. }

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/zzandliz/p/5023164.html