1655: [Usaco2006 Jan] Dollar Dayz 奶牛商店

1655: [Usaco2006 Jan] Dollar Dayz 奶牛商店

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 477  Solved: 253
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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).

    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

Source

Silver

背包水题。

f[i][j]表示用前j-1个数分解i的方案数，则：f[i][j]=f[i][j-1],f[i-j*u][j-1]。

显然第二维可以去掉。。

数据大，加个加法高精度就好了。

附代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<climits>#include<cstdlib>#include<cstring>#include<string>#include<fstream>#include<queue>#include<map>#include<set>#include<stack>#define N 1001using namespace std;int n,k;int f[N];struct node{int a[50],len;}p[N];void add(node &a,node &b){node c;int lenc=0;for(int i=0;i<=49;i++)c.a[i]=0;for(int i=1;i<=min(a.len,b.len);i++){c.a[i]+=a.a[i]+b.a[i];if(c.a[i]>9)c.a[i]-=10,c.a[i+1]++;}int j=min(a.len,b.len)+1;lenc=j-1;while(j<=a.len){c.a[j]+=a.a[j];if(c.a[j]>9)c.a[j]-=10,c.a[j+1]++;j++;lenc=a.len;}while(j<=b.len){c.a[j]+=b.a[j];if(c.a[j]>9)c.a[j]-=10,c.a[j+1]++;j++;lenc=b.len;}if(c.a[j])lenc++;c.len=lenc;a=c;}int main(){//freopen("in.txt","r",stdin);//freopen("my.txt","w",stdout);p[0].len=1,p[0].a[1]=1;for(int i=2;i<=49;i++)p[0].a[i]=0;scanf("%d%d",&n,&k);for(int j=1;j<=k;j++)for(int i=n;i>=j;i--)for(int u=1;u*j<=i;u++)add(p[i],p[i-u*j]);for(int i=p[n].len;i;i--)printf("%d",p[n].a[i]);}