Codeforces 552B - Vanya and Books(数学)

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本文介绍了一种计算从1到n所有整数所占用总位数的方法。通过分析不同位数区间的数值特点，利用数学公式高效求解，并提供了一份完整的C++实现代码。

B. Vanya and Books
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the n books should be assigned with a number from 1 to n. Naturally, distinct books should be assigned distinct numbers.

Vanya wants to know how many digits he will have to write down as he labels the books.

Input
The first line contains integer n (1 ≤ n ≤ 109) — the number of books in the library.

Output
Print the number of digits needed to number all the books.

Examples
input
13
output
17
input
4
output
4
Note
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.

Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits.

题意:
给出一个n,输出从1-n的所有数,一共要输出几个数字.
(比如17这个数要输出2个数字).

解题思路:
1-9有9个数
10-99有90个数
100-999有900个数
那么对于10^digit - 10(digit+1)-1 一共有9*10^digit个数.

对于最大长度的数出现的次数为 (n - pow(10, dight-1) + 1) * dight
剩下的累加计算即可.

AC代码:

#include <bits/stdc++.h>
using namespace std;
#define int long long
int pow(int a, int b)
{
    int tmp = a;
    for(int i = 1; i <= b; i++)
        a *= tmp;
    return a;
}
main()
{
    int n;
    cin >> n;
    int tmp = n;
    int dight = 0;
    while(tmp)
        dight++, tmp /= 10;
    int sum = 0;
    for(int i = 1;i <= dight-1; i++)
        sum += 9 * pow(10, i-1) * i;
    sum += (n - pow(10, dight-1) + 1) * dight;
    cout << sum << endl;
    return 0;
}