1049. Counting Ones (30)

Posted 2020-10-18 给杰瑞一块奶酪~

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


如果穷举的话太麻烦了，还要排着计算每个数有几个1。可以假定某一位是1，看一下此时有多少个数满足不超过n，假如32105这个数，假定十位是1，他的高位是321，低位是5，高位肯定不能是321（32110 > 32105），当高位取0-320时，低位可以取0-9，此时是（320 + 1）* 10。接洗下来假设百位是1，百位本来就是1，高位为32，低位为05，高位可以取0-32，高位取0-31时，低位可以取0-9，高位取32时，低位只可以取0-5，所以此时是32 * 100 + 5 + 1。继续假设千位是1，高位和低位都可以尽情取所在范围内的数，即高位取0-3，低位可以取0-999，结果很清楚了已经（3+1）*1000。可见只需要分所假设位是0,1，和其他的情况即可。
代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <set>
#include <cstring>
using namespace std;

int main()
{
    int n;
    cin>>n;
    int base = 1;
    int low,now,high;
    int ans = 0;
    while(n/base)
    {
        high = n / (base * 10);
        now = (n / base) % 10;
        low = n % base;
        if(now == 0)//now 位 要是1 则high必然要减小1
        ans += high * base;
        else if(now == 1)//now位本来就是1 则high位不变
        ans += high * base + low + 1;
        else//那么low可以取base个
        ans += (high + 1) * base;
        base *= 10;
    }
    cout<<ans;
}