coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

Source

HDU2010省赛集训队选拔赛（校内赛）

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题意：给你P a1 a5 a10 a50 a100 ，ai代表i角的硬币有ai个，问凑成P最少，最多的硬币个数。

题解：

最少直接从大到小取，最大就是不断的把面值大的转换成小的。渣渣只会暴力。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

#define N 1010
#define ll long long

using namespace std;

int s;
int a[7];///面值为c[i]的硬币剩余的个数为a[i]个
int b[7];///面值为c[i]的硬币用了b[i]个
int c[7]= {0,1,5,10,50,100};///面值

int main() {
    //freopen("in.txt","r",stdin);
    int t;
    cin>>t;
    while(t--) {
        scanf("%d",&s);
        for(int i=1; i<=5; i++) {
            scanf("%d",&a[i]);
        }
        int x=s;
        int Min=0,Max=0;
        memset(b,0,sizeof b);
        for(int i=5; i>=1; i--) {
            if(x>=c[i]) {
                int num=x/c[i];
                if(num>=a[i])
                    num=a[i];
                Min+=num;
                a[i]-=num;
                b[i]=num;
                x-=num*c[i];
            }
        }
        if(x!=0)Min=-1;
        if(Min==-1) {
            printf("-1 -1\n");
            continue;
        }
        Max=Min;
        ///求最大值就是大的尽量转换成小的
        while(1) {
            int flag=1; ///标记是否有转换
            for(int i=5; i>1; i--) {
                if(b[i]) {///用的个数
                    for(int j=i-1; j>=1; j--) {
                        if(b[i]==0)
                            break;
                            //转换
                        if(a[j]*c[j]>=c[i]) {
                            int x=a[j]*c[j]/c[i];
                            if(x>=b[i]) {
                                Max=Max-b[i]+c[i]*b[i]/c[j];
                                b[j]+=c[i]*b[i]/c[j];
                                a[i]+=b[i];
                                a[j]=a[j]-c[i]*b[i]/c[j];
                                b[i]=0;
                                flag=0;
                            } else {
                                Max=Max-x+c[i]*x/c[j];
                                b[j]+=c[i]*x/c[j];
                                a[i]+=x;
                                a[j]=a[j]-c[i]*x/c[j];
                                b[i]-=x;
                                flag=0;
                            }
                        }
                    }
                }
            }
            if(flag)
               break;
        }
        printf("%d %d\n",Min,Max);
    }
    return 0;
}