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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 (<=2^30^).
Output Specification:
For each test case, print the number of 1's in one line.
Sample Input:
12
Sample Output:
5
思路:
给出一个数字n,求1~n的所有数字里面出现1的个数
设当前处理至第k位,记left为第k位的高位所表示的数,now为第k位的数,right为第k位的低位所表示的数。
- 若now==0,则ans+=left*a;
- 若now==1,则ans+=left*a+right+1
- 若now>=2,则ans+=(left+1)*a
C++:
#include<cstdio>
int main(){
int n,a=1,ans=0;
int left,now,right;
scanf("%d",&n);
while(n/a!=0){
left=n/(a*10);
now=n/a%10;
right=n%a;
if(now==0)ans+=left*a;
else if(now==1)ans+=left*a+right+1;
else ans+=(left+1)*a;
a*=10;
}
printf("%d\n",ans);
return 0;
}