1049 counting ones

Posted 2020-06-18

从所有一位数包含的1开始向上递推所有k位数包含的1，递推式：

ak = ak-1 * 10 + pow(10,k-1);

AC代码：

#include <vector>
#include <cstdio>
using namespace std;
int main(){
    int n;
    vector<int> rec;
    scanf("%d",&n);
    int nc(n);
    while(n != 0){
        rec.push_back(n % 10);
        n /= 10;
    }
    vector<int> r(rec.size());
    vector<int> rem;
    int radix(1);
    for(int i = 0;i < r.size();i++){
        rem.push_back(radix);        
        if(i == 0)
            r[0] = 1;
        else{
            r[i] = r[i - 1] * 10 + radix;
        }
        radix *= 10;
    }
    int ret(0);
    for(int i = rec.size() - 1;i >= 0;i--){
        if(i != 0 && rec[i] != 0){
            if(rec[i] == 1)
                ret += r[i - 1] + nc % rem[i] + 1;
            else
                ret += r[i - 1] * rec[i] + rem[i];
        }
        else if(i == 0 && rec[i] != 0)
            ret += 1;
    }
    printf("%d\n",ret);
    return 0;
}