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．
Sample Input
5 3
Sample Output
5
HINT
f[j]=f[j]+f[j-i]，一个非常明显的方程。
但是此题需要高精度！
作为粗心人群就直接ull提交了，然后光荣WA
发现要高精度，
但是高精度预估了30位数……
发现极限数据非常长！！却直接submit了！！
吐血……要开40位才够。。
下次代码慢点打。。
#include<bits/stdc++.h>
using namespace std;
int f[1005][50];
void Plus(int x,int y){
int len=max(f[x][0],f[y][0]);
for (int i=1;i<=len;i++) f[x][i]=f[x][i]+f[y][i];
int i=1;
while (i<=len){
f[x][i+1]+=f[x][i]/10;
f[x][i]%=10;
if (f[x][len+1]) len++;
i++;
}
f[x][0]=len;
}
int main(){
int n,K;cin>>n>>K;
f[0][0]=f[0][1]=1;
for (int i=1;i<=K;i++)
for (int j=i;j<=n;j++) Plus(j,j-i);
for (int i=f[n][0];i;i--) printf("%d",f[n][i]);
putchar('\n');
return 0;
}
