## bzoj1655 [Usaco2006 Jan] Dollar Dayz 奶牛商店

2018年01月13日 ⁄ 综合 ⁄ 共 1502字 ⁄ 字号 评论关闭

## 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).

约翰到奶牛商场里买工具．商场里有K(1≤K≤100).种工具，价格分别为1，2，…，K美元．约翰手里有N(1≤N≤1000)美元，必须花完．那他有多少种购买的组合呢？

## Input

A single line with two space-separated integers: N and K.

仅一行，输入N，K.

## Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

不同的购买组合数．

5 3

## Sample Output

5

f[j]=f[j]+f[j - i]

```#include<cstdio>
struct bignum{
int len;
int a[100];
}f[1001];
int n,k;
inline int max(int a,int b)
{return a>b?a:b;}
inline int min(int a,int b)
{return a<b?a:b;}
inline void add(int i,int j,bignum &k)
{
bignum ans;
ans.len=max(f[j-i].len,f[j].len);
for (int l=1;l<=ans.len;l++)
ans.a[l]=f[j-i].a[l]+f[j].a[l];
ans.a[ans.len+1]=0;
for (int l=1;l<=ans.len;l++)
{
ans.a[l+1]+=ans.a[l]/10;
ans.a[l]%=10;
}
if (ans.a[ans.len+1])ans.len++;
k.len=ans.len;
for (int i=1;i<=ans.len;i++)
k.a[i]=ans.a[i];
}
int main()
{
scanf("%d%d",&n,&k);
f[0].len=f[0].a[1]=1;
for (int i=1;i<=min(n,k);i++)
for (int j=i;j<=n;j++)