poj 3181 动态规划

Dollar Dayz

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4840 Accepted: 1842

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

Source

USACO 2006 January Silver

给定n和k，k代表有k个商品价格分别是1-k，每个商品的数量无限，n代表有n元钱，求把这n元钱花完总共有多少种方法。

开始发现这题跟以前做的题很像，很快就写出了推导式，但是初始化的时候却出现了问题。最开始的时候将从1-k的dp全部设置为1，最后发现重复计算了很多。

后来网上搜了一下才明白要把dp【0】设置为1.这样就不会重复计算。

中间运算会超过long long的表示范围，因此用数组模拟就行。

dp[i][0..50]表示第i个数有多少种组合方式。dp数组第二维表示组合方式的数量。

#include<map>#include<vector>#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<stack>#include<queue>#include<set>#define inf 0x3f3f3f3f#define mem(a,x) memset(a,x,sizeof(a))using namespace std;typedef long long ll;typedef pair<int,int> pii;inline int in(){    int res=0;char c;    while((c=getchar())<'0' || c>'9');    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();    return res;}int dp[1111][55];int v[111];void add(int n,int m){    int tmp=0;    for(int i=0;i<50;i++)    {       dp[n][i]+=dp[m][i]+tmp;       if(dp[n][i]>=10)       {           tmp=1;           dp[n][i]%=10;       }       else tmp=0;    }}int main(){    int n,k;    while(~scanf("%d%d",&n,&k))    {        mem(dp,0);        dp[0][0]=1;        for(int i=1;i<=k;i++) //从每一个商品开始看它能组成哪些钱数        {            for(int j=i;j<=n;j++)            {                add(j,j-i);            }        }       int i=50;       while(i>0 && dp[n][i]==0)--i;       while(i>=0)       {           printf("%d",dp[n][i]);           --i;       }       puts("");    }    return 0;}