PTA 1049 Counting Ones

题目描述

分析：找出1~N之间的所有数字中1出现的次数，暴力会超时（但是考试时也会得到部分分）

#include<iostream>
#include<algorithm>
using namespace std;
int ans;
int main()
{
    int n;
    scanf("%d", &n);
    int a = 1, left, right;
    while (n / a)
    {
        int now = n / a % 10;
        left = n / (a * 10);
        right = n%a;
        if (now == 0)ans += left*a;
        else if (now == 1)ans += left*a + 1 + right;
        else if (now > 1)ans += (left + 1)*a;
        a *= 10;
        cin.get();
    }
    printf("%d", ans);
    return 0;
}