NEFU 2016省赛演练一 B题（递推）

HK

Problem:B

Time Limit:2000ms

Memory Limit:65535K

Description

yy is interested in numbers and yy numbers all of the day, Now yy give us the definition of the HK of a decimal number: if(0 <= x < 10)  HK(x) = x, else HK(bit[1]bit[2]..bit[nBit]) = HK(bit[1] + bit[2] +...+bit[nBit]), such as HK(2) = 2, HK(364) = HK(3 + 6 + 4) = HK(13) = HK(1 + 3) = HK(4) = 4. As you can see is easy to a decimal number X's HK(X),  but now yy want to know the smallest x such in the range of 1 to x (1 and x included, x >= 1)  that there exist n decimal numbers who's HK(x) = m.

Input

There are multi case end with EOF
Each  case has two numbers n, m as described above (0 <= n < 10^10000, 1<= m <= 9) 

Output

Every case print like this "Case #cas: k" k is the casenumber, k is the smallest x mod 1000000007. If there is no answer just let k = -1.

Sample Input

1 1
2 2
2 1

Sample Output

Case #1: 1
Case #2: 11
Case #3: 10

Hint

In the second case in the range of [1, 11] Only HK(2) and HK(11) equal 2 . So 11 is the smallest number
In the third case in the range of[1, 10] Only HK(1) and HK(10) equal 1. 10 is the smallest number

题意：如果(0 <= x < 10)  HK(x) = x, 否则 HK(bit[1]bit[2]..bit[nBit]) = HK(bit[1] + bit[2] +...+bit[nBit]), 
　　　例如 HK(2) = 2, HK(364) = HK(3 + 6 + 4) = HK(13) = HK(1 + 3) = HK(4) = 4.
　　　给定n和m，就是求第n个和HK（m）的数组下标。
题解：打表发现HK数组是123456789 123456789 123456789.............
　　　得出递归公式 ans=(n-1)*9+m;
　　　注意n高精度取余，用公式取余一下就行了。
　　　注意n=0的时候即存在第0个HK（m）的最小的数组下标，即ans的最小值也就是x最小值1。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#define mod 1000000007
using namespace std;
typedef long long ll;
char c[100005];
int main()
{
    ll i,n,m,cas=1;
    while(scanf("%s",c)!=EOF)
    {
        getchar();
        scanf("%lld",&m);
        int len=strlen(c);
        n=c[0]-'0';
        if(n==0)
        {
            printf("Case #%lld: 1
",cas++);
            continue;
        }
        for(i=1;i<len;i++)
        n=(n*10+c[i]-'0')%mod;
        ll ans=(9*n-9+m+mod)%mod;
        printf("Case #%lld: %lld
",cas++,ans);  //注意cas是long long 要用%lld
    }
    return 0;
}