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
Sample Output
就是很水的完全背包
f[j]=f[j]+f[j - i]
一开始以为是部分背包，然后样例都过不了……
改成完全背包又wa了……
结果发现USACO数据里有20多位的答案……
没办法改高精度吧……
#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++)
add(i,j,f[j]);
for (int i=f[n].len;i>=1;i--)
printf("%d",f[n].a[i]);
}