pat1049. Counting Ones (30)

1049. Counting Ones (30)

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

思路：

统计每位的1的贡献。

对于k位（k>=1）：

1.Ak=0，count+=AnAn-1....Ak+1AkAk-1....A1*10^(k-1)

2.Ak=1，count+=AnAn-1....Ak+1AkAk-1....A1*10^(k-1)+Ak-1Ak-2...A1+1

3.Ak>=2，count+=(AnAn-1....Ak+1AkAk-1....A1+1)*10^(k-1)

 #include<cstdio>
 #include<stack>
 #include<cstring>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 using namespace std;
 //count的最大值是1036019223
 int main(){
     int n;
     scanf("%d",&n);
     long long base=;
     long long count=;
     int frpart,afpart,a;
     while(n>=base){
         a=n/base%;
         frpart=n/(*base);
         afpart=n%base;
         count+=frpart*base;
         if(a==){
             count+=afpart+;
         }
         else if(a>){
             count+=base;
         }
         base*=;
     }
     printf("%lld\n",count);
     return ;
 }