Problem F

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

题意：有1 5 10 50 100这几种面值的钱，且个数是给定的，给你一个钱数，让你求最多硬币树，和最少硬币数；

解题思路：求最小钱数当然好求，从大到小开始贪心，但是求最多硬币数的时候就有点问题了，试了很多种方法，可以用小硬币来去换已经球出来的最小硬币的个数，还完了可能就是最大的了；

感悟：这两天做题有点慢，每个题都得先先想，而且还有很多没见过的东西，还得先查资料，得加快进度了。

代码：

#include
#include
#include
using namespace std;
int mina(int a[],int p,int a_m[])
{
    int ans=0;
    for(int i=5;i>1;i--)
    {
        if(p>=a[i]*a_m[i])//看看当前剩下的钱是不是比枚举到的钱数大
        {
            ans+=a[i];
            p-=a[i]*a_m[i];
        }
        else//小
        {
            ans+=p/a_m[i];
            p%=a_m[i];
        }

    }
    if(p>a[1]) return -1;
    else return ans+p;
}
int maxa(int a[],int p,int a_m[],int sum[])
{
    int ans=0;
    for(int i=5;i>1;i--)
    {
        if(p<=sum[i-1])//p比当前金币的总价值小
            continue;
        else//大
        {
            int t;
            t=((p-sum[i-1])/a_m[i])+(((p-sum[i-1])%a_m[i])?1:0);
            //(p-sum[i-1])是除去当前钱数之前的钱数之和
            ans+=t;
            p-=t*a_m[i];
            //printf("t=%d a[i]_m=%d p=%d\n",t,a_m[i],p);
        }
    }
    //printf("P=%d a[1]=%d\n",p,a[1]);
    if(p>a[1]) return -1;
    else return ans+p;

}
int solve(int a[],int p,int a_m[])
{
    int minn=0,maxn=0;
    int sum[6]={0};
    for(int i=1;i<6;i++)
        sum[i]=sum[i-1]+a[i]*a_m[i];
    minn=mina(a,p,a_m);
    maxn=maxa(a,p,a_m,sum);
    //printf("minn=%d maxn=%d\n",minn,maxn);
    if(minn==-1)printf("-1 -1\n");
    else
    {
        if(maxn==-1)printf("-1 -1\n");
        else
            printf("%d %d\n",minn,maxn);
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    int a[6],a_m[6]={0,1,5,10,50,100},sum=0,n,p;
    scanf("%d",&n);
    for(int i=0;i
    {
        memset(a,0,sizeof(a));
        scanf("%ld%ld%ld%ld%ld%ld",&p,&a[1],&a[2],&a[3],&a[4],&a[5]);
        for(int j=1;j<6;j++)
        {
            sum+=a_m[j]*a[j];
        }
        if(sum

        else solve(a,p,a_m);
    }
    return 0;
}

转载于:https://www.cnblogs.com/wuwangchuxin0924/p/5781663.html