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。

#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int n;
string s;
int getstr(string s){
    int num = 0;
    for(int i = 0; i < s.length(); i++)
        num = num*10 + s[i]-'0';
    return num;
}

ll pow(int x){
    int num = 1;
    while(x--){
        num *= 10;
    }
    return num;
}
int main(){
    cin >> s;
    int len = s.length();
    n = getstr(s);
    ll sum = 0;
    ll mod = pow(len-1), left = 0, right = 0;
    for(int i = 0; i < len; i++){
        right = n%mod;
        if(s[i] == '1'){
            sum += (right+1);
            sum += left*mod;
        }else if(s[i] == '0'){
            sum += left*mod;
        }else{
            sum += (left+1)*mod;
        }
        left = left*10 + s[i]-'0';
        mod /= 10;
    }
    cout << sum << endl;
    return 0;
}