Coins(多重背包)
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http://acm.hdu.edu.cn/diy/diy_previewproblem.php?cid=19572&pid=1017
Coins
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 2 Accepted Submission(s) : 1
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Problem Description
Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse 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
2009 Multi-University Training Contest 3 - Host by WHU
解析:
题意:给你n中硬币,及其数量,以及最大价值m
求在不超过m的情况下,从1到m可以构成多少种价值
思路:其实就是整数构造法,可以利用多重背包法进行处理,然后根据多重背包的定义从1枚举到m,求得最终结果
*/
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int maxn=100000+10;int f[maxn];int a[105],c[105];int n,m;int max(int a,int b){ return a>b? a:b;}void completepack(int cost){ for(int i=cost;i<=m;i++) f[i]=max(f[i],f[i-cost]+cost); //printf("f[1m]=%d\n",f[m]);}void zeroonepack(int cost){ for(int i=m;i>=cost;i--) {f[i]=max(f[i],f[i-cost]+cost); //printf("f[%d]=%d,cost=%d,\n",i,f[i],cost); }}void multipack(int cost,int amount){ if(cost*amount>=m) { completepack(cost); return ; } int k; k=1; while(k<amount) { zeroonepack(k*cost); amount=amount-k; k=k*2; } zeroonepack(amount*cost);}int main(){ int i,j; int ans; while(scanf("%d%d",&n,&m)!=EOF) { if(n==0&&m==0) break; for(i=0;i<n;i++) scanf("%d",&a[i]); for(j=0;j<n;j++) scanf("%d",&c[j]); memset(f,0,sizeof(f)); for(i=0;i<n;i++) {multipack(a[i],c[i]); // printf("a[i]=%d,c[i]=%d",a[i],c[i]); } ans=0; for(i=1;i<=m;i++) if(f[i]==i) { ans++; } printf("%d\n",ans); }}#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int maxn=100000+10;int f[maxn];int a[105],c[105];int n,m;int max(int a,int b){ return a>b? a:b;}void completepack(int cost){ for(int i=cost;i<=m;i++) f[i]=max(f[i],f[i-cost]+cost); //printf("f[1m]=%d\n",f[m]);}void zeroonepack(int cost){ for(int i=m;i>=cost;i--) {f[i]=max(f[i],f[i-cost]+cost); //printf("f[%d]=%d,cost=%d,\n",i,f[i],cost); }}void multipack(int cost,int amount)//多重背包{ if(cost*amount>=m) { completepack(cost); return ; } int k; k=1; while(k<amount) { zeroonepack(k*cost); amount=amount-k; k=k*2; } zeroonepack(amount*cost);}int main(){ int i,j; int ans; while(scanf("%d%d",&n,&m)!=EOF) { if(n==0&&m==0) break; for(i=0;i<n;i++) scanf("%d",&a[i]); for(j=0;j<n;j++) scanf("%d",&c[j]); memset(f,0,sizeof(f)); for(i=0;i<n;i++) {multipack(a[i],c[i]); // printf("a[i]=%d,c[i]=%d",a[i],c[i]); } ans=0; for(i=1;i<=m;i++)//利用多重背包,求得在体积1....m之间的最大价值 if(f[i]==i)//最大价值与所需的体积相同则说明可以构成该数 { ans++; } printf("%d\n",ans); }}
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