POJ1742 Coin [DP補完計劃]
阿新 • • 發佈：2018-05-17
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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.
題目傳送門
Coins| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 41707 | Accepted: 14125 |
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
分析：很顯然的多重背包，由數據範圍可知需要進行拆分優化或者用單調隊列優化。這裏蒟蒻還不會單調隊列優化，所以用的是二進制拆分。
Code：
1 //It is made by HolseLee on 17th May 2018 2 //POJ 1742 3 #include<cstdio> 4#include<cstring> 5 #include<cstdlib> 6 #include<cmath> 7 #include<iostream> 8 #include<iomanip> 9 #include<algorithm> 10 #define Fi(i,a,b) for(int i=a;i<=b;i++) 11 #define Fx(i,a,b) for(int i=a;i>=b;i--) 12 using namespace std; 13 const int N=1e5+7;bool dp[N]; 14 int n,m,cnt,ans,k[N],a[101],c[101]; 15 int main() 16 { 17 ios::sync_with_stdio(false); 18 while(cin>>n>>m){ 19 if(!n||!m)break; 20 memset(dp,false,sizeof(dp)); 21 dp[0]=true;cnt=0;ans=0; 22 Fi(i,1,n)cin>>a[i];Fi(i,1,n)cin>>c[i]; 23 Fi(i,1,n){ 24 for(int j=1;j<=c[i];j<<=1){ 25 k[++cnt]=a[i]*j;c[i]-=j;} 26 if(c[i])k[++cnt]=a[i]*c[i];} 27 Fi(i,1,cnt)Fx(j,m,k[i]){ 28 dp[j]|=dp[j-k[i]];} 29 Fi(i,1,m)if(dp[i])ans++; 30 cout<<ans<<"\n";} 31 return 0; 32 }
POJ1742 Coin [DP補完計劃]