1049 Counting Ones (30 分)

1049 Counting Ones (30 分)

 

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 (≤).

Output Specification:

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

Sample Input:

12

Sample Output:

5

30分的题确实比其他分的好很多，这种题就比较锻炼思维。
找1的个数，这个是一定不能暴力的，肯定超时。
可能就只能骗分。

 1 #include <bits/stdc++.h>
 2 #define ll long long int
 3 using namespace std;
 4 int n;
 5 string s;
 6 int getstr(string s){
 7     int num = 0;
 8     for(int i = 0; i < s.length(); i++)
 9         num = num*10 + s[i]-'0';
10     return num;
11 }
12 
13 ll pow(int x){
14     int num = 1;
15     while(x--){
16         num *= 10;
17     }
18     return num;
19 }
20 int main(){
21     cin >> s;
22     int len = s.length();
23     n = getstr(s);
24     ll sum = 0;
25     ll mod = pow(len-1), left = 0, right = 0;
26     for(int i = 0; i < len; i++){
27         right = n%mod;
28         if(s[i] == '1'){
29             sum += (right+1);
30             sum += left*mod;
31         }else if(s[i] == '0'){
32             sum += left*mod;
33         }else{
34             sum += (left+1)*mod;
35         }
36         left = left*10 + s[i]-'0';
37         mod /= 10;
38     }
39     cout << sum << endl;
40     return 0;
41 }