PAT--1049. Counting Ones

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

题解

显然，暴力~不可取，但可以用来对拍。

对于

1~n

这n个数，要知道数字1出现的次数，可以计算出个位、十位、百位…这些数位上1出现的次数，累加起来即可。

个位：0~9的循环，即0123456789，一次循环只有1个1

十位：0~9的循环，每个数连续出现10次, 即000000000111111111….一次循环有10个1

百位：0~9的循环，每个数连续出现100次，….一次循环有100个1

…

规律性是很显然的。

#include <bits/stdc++.h>
using namespace std;

int n, sum;

int length(int n){
int ret = 0;
while(n){
ret++;
n /= 10;
}
return ret;
}

int main(){
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif

cin >> n;
n++;

int k = 10;
int len = length(n);
for(int i = 1; i < len; ++i){
int a = (n / k * (k / 10));
int b = (n % k > k / 10) ? (n % k < 2 * k / 10 ? (n % k - k / 10) % (k / 10) : k / 10) : 0;
//cout << "a + b: " << a + b << endl;
//cout << "b: " << b << endl;
sum += a + b;
k *= 10;
}
// 处理最高位
int p = n / ((int)pow(10, len - 1));
if(p == 1) sum += n % (int)pow(10, len - 1);
else sum += pow(10, len - 1);

cout << sum << endl;

return 0;
}