CF 2014 ACM-ICPC Vietnam National Second Round C. ATM withdrawal

C. ATM withdrawal

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Vinh works for an ATM machine manufacturing company. The basic functionality of an ATM machine is cash withdrawal. When a user requests a cash withdrawal ofW VND (Vietnamese Dong), the ATM has to dispenseN
money notes such that they sum up to W. For the next generation of ATM machine, Vinh is working on an algorithm to minimize the numberN of money notes for each cash withdrawal transaction.

Your task is to help Vinh to do his job given that the money notes come in the values of1000, 2000, 3000, 5000, 1000 * 101, 2000 * 101, 3000 * 101, 5000 * 101, ..., 1000 * 10c, 2000 * 10c, 3000 * 10c, 5000 * 10c
where c is a positive integer and Vinh has unlimited supply of money notes for each value.

Input
The input file consists of several datasets. The first line of the input file contains the number of datasets which is a positive integer and is not greater than1000. The following lines describe the datasets.

The first line consists of one positive integer W(W ≤ 1018);
The second line consists of one positive integer c(c ≤ 15).

Output
For each dataset, write in one line two space-separated integers
N and S where
S is the number of ways to dispense the fewest number
N of money notes. In case there is no way to serve the cash withdrawal request, write out 0 in one line instead.

Sample test(s)

Input

2
1000
1
7000
1

Output

1 1
2 1

解析

官方题解

我先把数据缩小1000倍。主要是要处理一个问题：如果你只有1,2,3,5要凑k元，打个表就有规律了。

k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

note1 1 1 2 1 2 2 2 3 2 3 3 3 3 3

way 1 1 1 2 1 2 1 1 3 1 2 1 1 3 1

#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;
#define INF 0x3f3f3f3f
LL kind[]={1,2,3,5},
way[]={1,1,1,1,2,1,2,1,1,3, 1,2},
note[]={0,1,1,1,2,1,2,2,2,3, 2,3};
LL W,C;
void find(LL x,LL &ans_note,LL &ans_way)
{
LL Note=INF,Way=1;
Note=(x-1)/5+1;if(x%5==4) Note++,Way=3;
if(x%5==1) Way=2;
if(x==1) Way=1; if(x==4) Way=2;
ans_note=Note; ans_way=Way;
}
void work()
{
LL ans_way=1,ans_note=0;
for(int i=0;i<C && W>0;i++)
{
int dig=W%10;
ans_way*=way[dig];
ans_note+=note[dig];
W=W/10;
}
if(W) {LL x,y;find(W,x,y);ans_note+=x;ans_way*=y;}
printf("%I64d %I64d\n",ans_note,ans_way);
}

int main()
{
int T; scanf("%d",&T);
for(int i=1;i<=T;i++)
{
scanf("%I64d%I64d",&W,&C);
if(W%1000) puts("0");
else
{
W=W/1000;
work();
}
}
return 0;
}