Dollar Dayz（DP）

最新推荐文章于 2021-09-03 11:11:35 发布

Techay

最新推荐文章于 2021-09-03 11:11:35 发布

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分类专栏：动规

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本文介绍了一个利用动态规划算法解决FJ在The Cow Store购买工具问题的实例，详细阐述了如何通过组合数学原理计算不同花费下购买工具的组合方式。包括输入输出规范、算法实现细节以及优化策略，旨在帮助读者理解和掌握动态规划在实际问题中的应用。

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Description
Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1,$ 2, and $3. Farmer John has exactly$ 5 to spend. He can buy 5 tools at $1 each or 1 tool at$ 3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

    1 @ US$3 + 1 @ US$2

1 @ US $3 + 2 @ US$ 1
1 @ US $2 + 3 @ US$ 1
2 @ US $2 + 1 @ US$ 1
5 @ US $1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$ K (1 <= K <= 100).
Input
A single line with two space-separated integers: N and K.
Output
A single line with a single integer that is the number of unique ways FJ can spend his money.
Sample Input
5 3
Sample Output
5

用1到k的硬币能组成n的方法总数。

简单的dp
dp[i] = dp[i] + dp[i - k];
但是这题爆ll，手动开两个ll，一个存高位，一个存低位就好了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const ll mod = 1e16;
ll dp[2][1005];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    ll n,k;
    scanf("%lld%lld",&n,&k);
    memset(dp,0,sizeof(dp));
    dp[0][0] = 1;
    for(ll i = 1;i <= k;i++)
    {
        for(ll j = i;j <= n;j++)
        {
            dp[0][j] = dp[0][j] + dp[0][j - i];
            dp[1][j] = dp[1][j] + dp[1][j - i] + dp[0][j]/mod;
            dp[0][j] = dp[0][j]%mod;
        }
    }
    if(dp[1][n])
        printf("%lld",dp[1][n]);
    printf("%lld\n",dp[0][n]);
    return 0;
}