#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int num[6],sum[6];
int mi(int num[],int a[],int count1)
{
    int sum=0;
    for(int i=5;i>1;i--)
    {
        if(count1>a[i]*num[i])
        {
            sum=sum+num[i];
            count1=count1-a[i]*num[i];
        }
        else
        {
            int temp=count1/a[i];
            sum=sum+temp;
            count1=count1-temp*a[i];
        }
    }
    if(count1>num[1])
        return -1;
    else
        return sum+count1;
}
int ma(int sum[],int num[],int a[],int count1)
{
    int sum1=0;
    for(int i=5;i>1;i--)
    {
        if(count1>sum[i-1])
        {
            int t=(count1-sum[i-1])/a[i];
            if((count1-sum[i-1])%a[i]>0)
                t++;
            sum1=sum1+t;
            count1=count1-t*a[i];
        }

}
    if(count1>num[1])
        return -1;
    else
        return sum1+count1;
}
void pd(int sum[],int num[],int a[],int count1)
{
    int tmin=mi(num,a,count1);
    if(tmin==-1)
        printf("-1 -1\n");
    else
    {
        int tmax=ma(sum,num,a,count1);
        if(tmax==-1)
            printf("-1 -1\n");
        else
            printf("%d %d\n",tmin,tmax);
    }
}
int main()
{
    int t;
  int  a[6]={0,1,5,10,50,100};
    scanf("%d",&t);
    while(t--)
    {

int count1;
        scanf("%d",&count1);
        for(int i=1;i<=5;i++)
        {
            scanf("%d",&num[i]);
        }
        for(int i=1;i<=5;i++)
            sum[i]=sum[i-1]+a[i]*num[i];

if(sum[5]<count1)
            printf("-1 -1\n");
        else
            pd(sum,num,a,count1);
    }
    return 0;
}

coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1100    Accepted Submission(s): 316

Problem Description

"Yakexi, this is the best age!" Dong MW works hard and get high pay, he has many 1 Jiao and 5 Jiao banknotes(纸币), some day he went to a bank and changes part of his money into 1 Yuan, 5 Yuan, 10 Yuan.(1 Yuan = 10 Jiao)
"Thanks to the best age, I can buy many things!" Now Dong MW has a book to buy, it costs P Jiao. He wonders how many banknotes at least,and how many banknotes at most he can use to buy this nice book. Dong MW is a bit strange, he doesn‘t like to get the change, that is, he will give the bookseller exactly P Jiao.

Input

T(T<=100) in the first line, indicating the case number.
T lines with 6 integers each:
P a1 a5 a10 a50 a100
ai means number of i-Jiao banknotes.
All integers are smaller than 1000000.

Output

Two integers A,B for each case, A is the fewest number of banknotes to buy the book exactly, and B is the largest number to buy exactly.If Dong MW can‘t buy the book with no change, output "-1 -1".

Sample Input

3 33 6 6 6 6 6 10 10 10 10 10 10 11 0 1 20 20 20

Sample Output

6 9 1 10 -1 -1

Author

madfrog

代码