poj1742Coins（多重背包）

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

做这个题时，第一次这么写的：
#include<iostream>
#include<cstdio>
using namespace std;
bool dp[100][100001];
int n,m,a[100],su[100];
int main(){
while(cin>>n>>m){
if(n==0&&m==0) return 0;
for(int i=0;i<n;i++)
for(int j=1;j<=m;j++)
dp[i][j]=false;
for(int i=0;i<n;i++)
scanf("%d",&a[i]),dp[i][0]=true; //dp[i][j]为真，表示j元可以凑出来 ,a[i]硬币的面额
for(int i=0;i<n;i++)
scanf("%d",&su[i]); //su[i]表示面额为a[i]的硬币的数量
for(int i=1;i<=su[0]&&a[0]*i<=m;i++) //注意a[0]*i<=m，or就会出现runtime error
dp[0][a[0]*i]=true;
for(int i=1;i<n;i++){
for(int j=1;j<=m;j++){
for(int k=1;k<=su[i];k++){
if(j-k*a[i]>=0&&dp[i-1][j-k*a[i]]) dp[i][j]=true;
else dp[i][j]=dp[i-1][j];
}
}
}
int s=0;
for(int i=1;i<=m;i++){
if(dp[n-1][i]) /*printf("%d ",i),*/s++;
}
printf("%d\n",s);
}
}
这么写的话会超时，虽然这个dp好理解，但是这个dp的复杂度为O(m(su[i]的和))
现在附上第二次AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[2][100005];
int main(){
int n,m,va[102],su[102];
while(cin>>n>>m){
if(n==0&&m==0) return 0;
memset(dp,-1,sizeof(dp));
for(int i=0;i<n;i++)
scanf("%d",&va[i]); //硬币的面额
for(int i=0;i<n;i++)
scanf("%d",&su[i]); //面额为va[i]的硬币的数量
for(int i=0;i<=su[0]&&va[0]*i<=m;i++)
dp[0][va[0]*i]=su[0]-i;
for(int i=1;i<n;i++){
for(int j=0;j<=m;j++){
if(dp[(i-1)%2][j]>=0) dp[i%2][j]=su[i];
else if(j<va[i]||dp[i%2][j-va[i]]<=0) dp[i%2][j]=-1;
else dp[i%2][j]=dp[i%2][j-va[i]]-1;
}
}
int s=0;
for(int i=1;i<=m;i++)
if(dp[(n-1)%2][i]>=0) /*printf("%d ",i),*/s++;
printf("%d\n",s);
}
}
这个复杂度为O(nm),
dp[i][j]表示前i+1中硬币加和得到j元钱时，第i+1种硬币剩余的数量。