poj1742 Coins Dynamic Planning 다중 가방 추가

Coins
Time Limit: 3000MS
 
Memory Limit: 30000K
Total Submissions: 35350
 
Accepted: 12015
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

Source
LouTiancheng@POJ
자세한 강의 내용은 도전 프로그램 설계 경연 p62~p64 참조
얼떨결에 A 를 했다
왜 j 잘 생각한 다음에 이 편을 보충해라코드: #include #include #include #include #include #include using namespace std; const int maxn = 1e5 + 10; int val[maxn]; int num[maxn]; int dp[maxn]; int main(){ int n,m,i,j; while(scanf("%d%d",&n,&m)){ if(n==0&&m==0){ break; } for(i=1;i<=n;i++){ scanf("%d",val+i); } for(i=1;i<=n;i++){ scanf("%d",num+i); } memset(dp, -1, sizeof(dp)); dp[0]=0; for(i=1;i<=n;i++){ for(j=0;j<=m;j++){ if(dp[j]>=0){ dp[j] = num[i]; }else if(j=0){ res++; } } printf("%d",res); } return 0; }에서