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=1;
    long long count=0;
    int frpart,afpart,a;
    while(n>=base){
        a=n/base%10;
        frpart=n/(10*base);
        afpart=n%base;
        count+=frpart*base;
        if(a==1){
            count+=afpart+1;
        }
        else if(a>1){
            count+=base;
        }
        base*=10;
    }
    printf("%lld
",count);
    return 0;
}