Difference HDU - 5936

最新推荐文章于 2024-07-05 13:57:19 发布

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最新推荐文章于 2024-07-05 13:57:19 发布

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Little Ruins is playing a number game, first he chooses two positive integers y y and K K and calculates f(y,K) f(y,K), here

f (y, K) = \sum z in every digits of y z K (f (233, 2) = 22 + 32 + 32 = 22)

then he gets the result

x = f (y, K) - y

As Ruins is forgetful, a few seconds later, he only remembers K K, x x and forgets y y. please help him find how many y y satisfy x=f(y,K)−y x=f(y,K)−y.

Input

First line contains an integer T T, which indicates the number of test cases.

Every test case contains one line with two integers x x, K K.

Limits
1≤T≤100 1≤T≤100
0≤x≤109 0≤x≤109
1≤K≤9 1≤K≤9

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result.

Sample Input

2
2 2
3 2

Sample Output

Case #1: 1
Case #2: 2
#include<iostream>
using namespace std;
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<map>
#define LL long long

const int e[]={1, 10, 100, 1000, 10000, 100000};

int mi[13][13];
map<LL,int>coun[10];
void init()
{
      for(int i=0;i<10;i++){
         mi[0][i]=1;
        for(int j=1;j<10;j++)mi[j][i]=mi[j-1][i]*i;
    }
    for(int i=1;i<10;i++)
    {
        for(int j=0;j<e[5];j++)
        { LL sum=-j;
            for(int t=0;t<=4;++t)
                sum+=mi[i][j/e[t]%10];
             coun[i][sum]++;
        }
    }
}
int main() {
     init();
     int T;
    scanf("%d",&T);
    for(int k=1; k<=T; k++) {
        int x,b;
        LL ans=0;
       scanf("%d%d",&x,&b);
        for(int i=0;i<100000;i++)
        {
            LL c,d;
            d=-(LL)i*100000;
             for(int j=0;j<=4;++j)
            d+=mi[b][i/e[j]%10];
            c=x-d;
            if(coun[b].find(c)!=coun[b].end())//这里画重点，不知道为什么加就对，不加就不对，
            ans=ans+coun[b][c];			//如果知道可以评论。
        }
        if(x==0)
            ans--;
        printf("Case #%d: %lld\n",k,ans);
    }
    return 0;
}