递归之dfs Exchange Cards

最新推荐文章于 2022-04-01 18:42:08 发布

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本文介绍了一个基于篮球卡片价值和数量的交换算法问题。该算法帮助计算玩家如何利用手中不同价值的卡片来等价交换目标卡片的方法数。输入包括目标卡片价值、玩家拥有的卡片种类及其价值和数量，输出为可行的交换方案总数。

Exchange Cards

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13 Accepted Submission(s) : 10

Problem Description

As a basketball fan, Mike is also fond of collecting basketball player cards. But as a student, he can not always get the money to buy new cards, so sometimes he will exchange with his friends for cards he likes. Of course, different cards have different value, and Mike must use cards he owns to get the new one. For example, to get a card of value 10$, he can use two 5$ cards or three 3$ cards plus one 1$ card, depending on the kinds of cards he have and the number of each kind of card. And Sometimes he will involve unfortunately in a bad condition that he has not got the exact value of the card he is looking for (fans always exchange cards for equivalent value).

Here comes the problem, given the card value he plans to get and the cards he has, Mike wants to fix how many ways he can get it. So it's you task to write a program to figure it out.

Input

The problem consists of multiple test cases, terminated by EOF. There's a blank line between two inputs.

The first line of each test case gives n, the value of the card Mike plans to get andm, the number of different kinds of cards Mike has. n will be an integer number between 1 and 1000.m will be an integer number between 1 and 10.

The next m lines give the information of different kinds of cards Mike have. Each line contains two integers,val and num, representing the value of this kind of card, and the number of this kind of card Mike have.

Note: different kinds of cards will have different value, eachval and num will be an integer greater than zero.

Output

For each test case, output in one line the number of different ways Mike could exchange for the card he wants. You can be sure that the output will fall into an integer value.

Output a blank line between two test cases.

Sample Input

Sample Output

1

7

Source

Zhejiang University Local Contest 2006

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

int n,m;

struct card
{
    int val;
    int num;
}p[15];

int cnt;

void dfs(int cur,int total,int tar)
{
    if(total==tar)
    {
        cnt++;
        return;
    }
    if(total>tar||cur==m)
    return;
    for(int i=0;i<=p[cur].num;i++)
    dfs(cur+1,total+i*p[cur].val,tar);
}

int main()
{
    bool flag=false;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(flag)
        printf("\n");
        for(int i=0;i<m;i++)
        scanf("%d%d",&p[i].val,&p[i].num);
        cnt=0;
        dfs(0,0,n);
        printf("%d\n",cnt);
        flag=true;
    }
    return 0;
}