课程练习一 Problem F id： 1005

最新推荐文章于 2025-05-05 15:59:10 发布

原创最新推荐文章于 2025-05-05 15:59:10 发布 · 562 阅读

0 ·

CC 4.0 BY-SA版权

本文探讨了一道关于使用不同面额纸币支付特定金额的问题。通过算法设计，实现了求解最少及最多纸币数量的方法，并附上了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

*题目 Problem F id： 1005

*题意：给你一个价格，还有面值分别为1,5,10,50,100（单位：毛）纸币的数量，要你用最少数量的纸币和最多数量的凑出这个价格，输出最少和最多的数量。

*解题思路：因为我们要求的是花的最多数量纸币，所以就是要保证手上的纸币数量最少！！这样想的话问题就比较简单了，就转化为最少数量问题了。假设手上总共有p毛，而价格为q毛，我们用手上最少的数量的纸币去凑（p-q）毛，然后再用总数量减去该最少数量即可。

*感想：放了几天假，忘了写了啥

*AC源码

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

# define N 1010
# define ll long long

using namespace std;

int s,i;
int a[7];
int b[7];
int c[7]= {0,1,5,10,50,100};

int main() {

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( 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;

while(1) {
int flag=1;
for(int i=5; i>1; i--) {
if(b[i]) {
for(int j=i-1; j>=1; j--) {
if(b[i]==0)
break;

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;
}

Problem F

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 161 Accepted Submission(s) : 59

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

Statistic | Submit | Back