題意：

給定n種面值的硬幣面值分別為$W_i$個數為$C_i$，問用這些硬幣可以組成1~m之間的多少面值。

題目：

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

分析：

1.關于是否裝滿問題，其實之前的博客討論過這個問題，但只需考慮裝滿問題時，就可以考慮情況，“標記當前出現狀況”的形式來計算裝滿問題。
2.定義一個sum數組。每次填dp[j]時直接由dp[j-a[i]]推出，前提是sum[j-a[i]]<b[i]。sum每填一行都要清零，sum[j]表示當前物品填充j大小的包需要至少使用多少個。復雜度O(n*m)。

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int M=1e5+10; int n,m,ans; int a[M],b[M],num[M],dp[M]; int main(){while(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;ans=0;memset(dp,0,sizeof(dp));dp[0]=1;for(int i=0;i<n;i++)scanf("%d",&a[i]);for(int i=0;i<n;i++)scanf("%d",&b[i]);for(int i=0;i<n;i++){memset(num,0,sizeof(num));for(int j=a[i];j<=m;j++){if(!dp[j]&&dp[j-a[i]]&&num[j-a[i]]<b[i]){dp[j]=1;ans++;num[j]=num[j-a[i]]+1;}}}printf("%d\n",ans);}return 0; }

總結

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