划分数+划分后大数处理 POJ 3181—Dollar Dayz（ 三种思想 层层递进）

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

1.我看到本题时， 首先想到的是硬币种类组合问题，类似uva 147

例如：5 3 即 用 1 2 3若干枚硬币组合出5

转移方程：dp[i]=dp[i]+dp[i-coin[j]]//coin[j]即某硬币

代码：

//此代码不详细解释 建议做做uva  147题目
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1010;
const int maxm=110;
int dp[maxn],a[maxm];
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		dp[0]=1;
		for(int i=1;i<=m;i++)
			a[i]=i;
		for(int i=1;i<=m;i++)
		{
			for(int j=a[i];j<=n;j++)
			{
				dp[j]+=dp[j-a[i]];
			}
		 } 
		printf("%d\n",dp[n]);
	}
	return 0;
 } 

其实本题未做完我就意识到问题了，输入的m<=100 即种类数太多，针对本题此dp有所欠缺，

但是毕竟过了样例，我还是厚着脸皮交了一发（手动滑稽）

2.硬币种类和问题过不了，后来经过分析可以看出本题属于划分数问题

样例：5 3 对应5的3划分

划分数转移方程：j>=i dp[i][j]=d[i-1][j]+dp[i][j-i]//j的i划分 j<i dp[i][j]=dp[i-1][j]

分析到这，屁颠屁颠敲好代码，过了样例（真高兴），后又提交一发。

代码：

//划分数基础模板，相关自行百度学习
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[1010][1010];
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		dp[0][0]=1; 
		for(int i=1;i<=m;i++)
		for(int j=0;j<=n;j++)
		{
			if(j>=i) dp[i][j]=dp[i-1][j]+dp[i][j-i];
			else dp[i][j]=dp[i-1][j];
		}
		printf("%d\n",dp[m][n]);
	}
	return 0;
} 

结果还是wa了，实在没办法，后来参考了网上相关代码，才发现划分数有时会比较大，甚至超过long long 范围，因此有些题目自带取余，而本题则需要用到高低位分开计算。

3.高低位分开计算：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=1e18;
ll high[1010][1005],low[1010][1005];
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		memset(high,0,sizeof(high));//高位 
		memset(low,0,sizeof(low));//低位 
		low[0][0]=1;//低位初值为1 
		for(int i=1;i<=m;i++)//同划分数写法 
		for(int j=0;j<=n;j++)
		{
			if(j>=i){
				//对低位 高位分别进行划分数动规 
				low[i][j]=low[i-1][j]+low[i][j-i]; 
				high[i][j]=high[i-1][j]+high[i][j-i]+low[i][j]/inf;//高位需要来自低位的进位 
				low[i][j]%=inf;//低位控制在inf以内 
			}
			else {
				low[i][j]=low[i-1][j];
				high[i][j]=high[i-1][j];
			}
		}
		if(high[m][n]) printf("%lld",high[m][n]);//如果高位存在 
		printf("%lld\n",low[m][n]);
	}
	return 0;
} 

最后，过了，嘻嘻嘻~