ZOJ3449 Doraemon's Number Game III

数根的扩展版，把一个十进制的数，写成按b进制形式的十进制数，一直下去，直到在[0,10)之间。

有a(n)*10^n+a(n-1)*10^(n-1)...+a(0) = a(n)*b^n+a(n-1)*b^(n-1)...+a(0) mod (10 - b)
然后答案就是(n - 10) % (10 - b) + b了。

#include <iostream>
#include <math.h>
#include <deque>
#include <string>
#include <vector>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;

const int MAX = 100005;

char ch[MAX];
int len;

int go(int b)
{
    if(len == 1 && ch[0] <= '9')  return ch[0] - '0';
    int res = 0;
    for(int i = 0; ch[i] != '\0'; i++)
    {
        res = res * 10 + ch[i] - '0';
        res %= (10 - b);
    }
    res = (res - 10) % (10 - b);
    while(res < 0)  res += (10 - b);
    return res + b;
}

int main()
{
    while(gets(ch))
    {
        len = strlen(ch);
        for(int i = 1; i <= 9; i++)
        {
            if(i == 1)  printf("%d", go(i));
            else  printf(" %d", go(i));
        }
        printf("\n");
    }
}