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
題意：
給出一個數，輸出比它小的所有數的所有位中數字1出現的次數
分析：
從所給數的最低位起按位計算
代碼：
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int N, a=1, ans=0;//a爲10的當前位數
int left, now, right;
cin>>N;
while(N/a!=0){
left = N/(a*10);//left爲當前位左邊的部分
now = (N/a)%10;//now爲當前位
right = N%a;//right爲當前位右邊位
if(now==0) ans += left*a;
//左邊從0變化到比現在小1,不管右邊是多少,當前位都可以取到1
else if(now==1) ans += left*a + right + 1;
//左邊從0到比現在小1乘右邊位數，加上左邊爲當前值加右邊值從0到現在值
else ans += (left+1)*a;//左邊從0到現在不管右邊取何值都可以
a *= 10;
}
cout<<ans;
return 0;
}