1049. Counting Ones (30)

1049. Counting Ones (30)

http://pat.zju.edu.cn/contests/pat-a-practise/1049

输入N，输出1到N之间有多少个1。譬如，1到13之间有1,10,11,12,13 6个数字1

思路：按位数，找规律。参考

例如100

个位1的个数：它左边只能取0-9，所以有10个

十位1的个数：它左边只能取0，右边能取0-9，所以有10个

百位1的个数：1个

#include <stdio.h>

int main(){
    int d;
    scanf("%d",&d);
    int x=1;
    int total=0;
    while(d/x!=0){
        int right = d%x;
        int left = d/(x*10);
        int current = (d/x)%10;
        if(current==0) 
            total +=left*x;
        else if(current==1)
            total += left*x+right+1;
        else
            total += (left+1)*x;
        x = x*10;
    }
    printf("%d\n",total);
    return 0;
}