1049 Counting Ones

最新推荐文章于 2026-01-08 22:28:34 发布

原创最新推荐文章于 2026-01-08 22:28:34 发布 · 154 阅读

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#PTA #算法

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该博客介绍了一种算法，用于计算从1到给定正整数N之间所有数字中1的总数。通过数学推算，分别计算每个位上1出现的次数，最后给出C/C++代码实现。

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:

Sample Output:

题目大意

求1~N出现了几个1

1~11算出现4个1，即 1，10，11

思路

数学题，可以直接推算出每个位数1出现多少次

C/C++

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
    LL N,op=1,result=0;
    cin >> N;
    while (N>=op){
        LL key = N/op%10;
        LL left =  N/(op*10);
        LL right = N%op;
        if(key==0) result += left*op;
        else if(key==1) result += left*op+right+1;
        else result += (left+1)*op;
        op*=10;
    }
    cout << result << endl;
    return 0;
}