A1049. 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:

Sample Output:

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int main(){
 5     long long N, a = 10, ans = 0, left, mid, right;
 6     scanf("%lld", &N);
 7     while(N / (a / 10) != 0){
 8         left = N / a;
 9         mid = N % a / (a / 10);
10         right = N % (a / 10);
11         if(mid == 0)
12             ans += left * (a / 10);
13         else if(mid == 1)
14             ans += left * (a / 10) + right + 1;
15         else if(mid > 1)
16             ans += (left + 1) * (a / 10);
17         a *= 10;
18     }
19     printf("%d", ans);
20     cin >> N;
21     return 0;
22 }

View Code

总结：

1、本题应寻找规律完成。对于数字abcde来说，讨论第c位为1的有多少个数字，分为左边：ab，中间c，右边de。当c = 1时，ab可以取0到ab的任意数字，当取0到ab - 1时，de位任意取。当ab取ab时，de位只能从0 到 de，故共有ab * 1000 + de + 1个。当c = 0时，为了让c能 = 1， ab位只能取0 到 ab - 1共ab种可能，de位任取，故共有ab*1000种可能。当c > 1时，ab位可以取0到ab共ab+1种可能，de任取，故共有（ab + 1）* 1000种可能。