Dollar Dayz POJ - 3181

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

详解：http://blog.csdn.net/libin56842/article/details/9455979
有n个无区别的物品，将它们划分成不超过m组，这样的划分被称作n的m划分，特别地，m=n时称作n的划分数。用dp[i][j]表示j的i划分的总数，dp[i][j]=dp[i][j-i]+dp[i-1][j],复杂度O(nm)

----------------------------摘自《挑战程序设计竞赛》

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
typedef  long long ll;

int n,k; 
ll a[1100],b[1100],INF;

void Init(){
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    INF=1;
    for(int i=0;i<18;i++) INF=INF*10;
}

void Solve(){
    a[0]=1;
    for(int i=1;i<=k;i++){
        for(int j=1;j<=n;j++){
            if(j-i<0) continue;
            b[j]=b[j]+b[j-i]+(a[j]+a[j-i])/INF;
            a[j]=(a[j]+a[j-i])%INF;
        }
    }
    
    if(b[n]) printf("%lld",b[n]);
    printf("%lld\n",a[n]);
}

int main()
{   cin>>n>>k;
    Init();
    Solve();
    return 0;
}