POJ_1742_Coins_(动态规划,多重部分和)

描述

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http://poj.org/problem?id=1742

n种不同面额的硬币 ai ,每种各 mi 个,判断可以从这些数字值中选出若干使它们组成的面额恰好为 k 的 k 的个数.

原型:

n种不同大小的数字 ai ,每种各 mi 个,判断是否可以从这些数字之中选出若干使它们的和恰好为 k .

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 33732		Accepted: 11453

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and
C1,C2,C3...Cn corresponding to the number of Tony's coins of value
A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay
use these coins.

Input

The
input contains several test cases. The first line of each test case
contains two integers n(1<=n<=100),m(m<=100000).The second line
contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn
(1<=Ai<=100000,1<=Ci<=1000). The last test case is followed
by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

Source

LouTiancheng@POJ

分析

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图文详解:

http://www.hankcs.com/program/cpp/poj-1742-coins.html

用f[j]表示在第i次循环时,用前i-1中硬币组成面额j,第i-1中硬币还剩多少个,如果不能组成面额j,则f[j]=-1.

最后统计f[j]>=0(1<=j<=m)的个数即可.

注意:

1.起始时f[0]=0,表示刚开始可以组成面额0,且之前啥也不剩.

2.之后每次j要从0开始循环,要更新组成面额0(也就是啥都不组成)之后,第i-1种硬币还剩多少个.

3.bind1st表示一元什么什么...也可以写成 " bind2nd(greater_equal<int>(),0) ",这东西貌似在<set>头文件里.

ps.前几天才看过书上的多重部分和问题,这题简直就是模板,然而几乎忘得一干二净了,走马观花真是啥用都没有= =只有真正想明白了才算掌握了一点吧.

 #include<cstdio>
 #include<cstring>
 #include<set>
 #include<algorithm>
 using namespace std;

 const int maxn=,maxm=;
 int n,m;
 int a[maxn],c[maxn],f[maxm];

 void solve()
 {
     memset(f,-,sizeof(f));
     f[]=;
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=m;j++)
         {
             if(f[j]>=)
             {
                 f[j]=c[i];
             }
             else if(a[i]>j||f[j-a[i]]<=)
             {
                 f[j]=-;
             }
             else
             {
                 f[j]=f[j-a[i]]-;
             }
         }
     }
     int ans=count_if(f+,f+m+,bind1st(less_equal<int>(),));//统计满足f[j]>=0的个数
     printf("%d\n",ans);
 }    

 void init()
 {
     while(scanf("%d%d",&n,&m)==&&(n!=||m!=))
     {
         for(int i=;i<=n;i++) scanf("%d",&a[i]);
         for(int i=;i<=n;i++) scanf("%d",&c[i]);
         solve();
     }
 }

 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("coin.in","r",stdin);
     freopen("coin.out","w",stdout);
 #endif
     init();
 #ifndef ONLINE_JUDEG
     fclose(stdin);
     fclose(stdout);
     system("coin.out");
 #endif
     return ;
 }