POJ3181Dollar Dayz 完全背包

易水人去，明月如霜。

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

5 3

Sample Output

5

题意：给出两个数,n,m,问m以内的整数有多少种组成n的方法

思路：

这个题目是个比较明显的动态规划，如果想不到是背包问题，也可以写出状态转移方程如下：

用a[i][j]表示考虑到用数j进行拼接时数字i的拼接方法，可以得到状态转移方程如下：

a[i][j]=a[i][j-1]+a[i-j][j-1]+a[i-2j][j-1]+a[i-3j][j-1]…+a[0][j-1]意思很明显，就将j-1状态可以到达a[i][j]的状态的数字相加

其实这个题有更快的方法，看上面这个式子a[i][j]=a[i][j-1]+a[i-j][j-1]+a[i-2j][j-1]+a[i-3j][j-1]…+a[0][j-1]我们可以发现，其实可以转到a[i][j]的状态有两种，一种是a[i][j-1]就是不用j这个数字拼接i这个数字的方法数，另一种是a[i-j][j]就是用了j这个数字拼接的到i-j的方法数那么状态转移方程就可以写成a[i][j]=a[i][j-1]+a[i-j][j]不用加那么多项

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;long long a[1100][110],b[1100][110],inf;int main(){    int n,k;    scanf("%d%d",&n,&k);    inf=1;    for(int i=1;i<=18;i++) inf*=10;    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    for(int i=0;i<=k;i++) a[0][i]=1;    for(int i=1;i<=k;i++)    {        for(int j=1;j<=n;j++)        {    if(j-i<0){                b[j][i]=b[j][i-1];                a[j][i]=a[j][i-1];                continue;            }            b[j][i]=b[j][i-1]+b[j-i][i]+(a[j][i-1]+a[j-i][i])/inf;            a[j][i]=(a[j][i-1]+a[j-i][i])%inf;        }    }    if(b[n][k]) printf("%lld",b[n][k]);    printf("%lld",a[n][k]); return 0;}