HDU-2844-Coins (Multi-pack), hdu-2844-coins backpack

HDU-2844-Coins (Multi-pack), hdu-2844-coins backpack
Problem DescriptionWhuacmers 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 wocould 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.
InputThe 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.
OutputFor each test case output the answer on a single line.
Sample Input

3 101 2 4 2 1 12 51 4 2 10 0

 
Sample Output

84

 
Source2009 Multi-University Training Contest 3-Host by WHU
Train of Thought: Typical knapsack problems.

#include <cstdio>#include <algorithm>using namespace std;int d[105],c[105],dp[100005];int main(){    int n,m,i,j,k,ans;    while(~scanf("%d%d",&n,&m) && n+m)    {        for(i=0;i<n;i++) scanf("%d",&d[i]);        for(i=0;i<n;i++) scanf("%d",&c[i]);        for(i=0;i<=m;i++) dp[i]=0;        for(i=0;i<n;i++)        {            if(d[i]*c[i]>=m)            {                for(j=d[i];j<=m;j++) dp[j]=max(dp[j],dp[j-d[i]]+d[i]);            }            else            {                k=1;                while(((k<<1)-1)<=c[i])                {                    for(j=m;j>=k*d[i];j--) dp[j]=max(dp[j],dp[j-k*d[i]]+k*d[i]);                    k<<=1;                }                if(k-1<c[i])                {                    k=(c[i]-k+1)*d[i];                    for(j=m;j>=k;j--) dp[j]=max(dp[j],dp[j-k]+k);                }            }        }        ans=0;        for(i=1;i<=m;i++) if(dp[i]==i) ans++;        printf("%d\n",ans);    }}

HDU 1059, multiple backpacks; I use parity to change the number of items to 0 and 1

The complexity of the algorithm is n * m.
# Include <stdio. h>
Const int MAX = 20000*6 + 5;
Struct
{
Bool visited;
Int s [6];
} Dp [MAX];
Int max [6];
Bool check ()
{
Int I;
For (I = 0; I <6; I ++)
If (max [I])
Return 0;
Return 1;
}
Int main ()
{
Int n, I, j, tmp, CS = 1, k;
While (scanf ("% d", & max [0])! = EOF)
{
N = max [0];
For (I = 1; I <6; I ++)
{
Scanf ("% d", & max [I]);
N + = (I + 1) * max [I];
}
If (check ())
Break;
Printf ("Collection # % d: \ n", CS ++ );
If (n % 2)
{
Puts ("Can't be divided .");
Puts ("");
Continue;
}
N> = 1;
For (I = 0; I <= n; I ++)
Dp [I]. visited = 0;
Dp [0]. visited = 1;
For (I = 0; I <6; I ++)
Dp [0]. s [I] = 0;
For (I = 0; I <6; I ++)
{
For (j = 0; j + I + 1 <= n; j ++)
{
If (dp [j]. visited = 0)
Continue;
If (dp [j]. s [I]> = max [I])
Continue;
If (dp [j + I + 1]. visited = 0)
{
Dp [j + I + 1]. visited = 1;
For (k = 0; k <6; k ++)
Dp [j + I + 1]. s [k] = dp [j]. s [k];
Dp [j + I + 1]. s [I] ++;
}
}
}
If (dp [n]. visited)
Puts ("Can be divided .");
Else
Puts ("Can't be divided... & quo ...... full text>

Multiple knapsack problems

The patient has chronic pharyngitis and must take medicine for treatment.