HDU2844: Coins (multiple backpacks)

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 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.
 

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
2 5
1 4 2 1
0 0

Sample Output
8
4
 

A simple template question.

 

# Include <stdio. h> # include <algorithm> # include <string. h> using namespace std; const int MAX = 100000; int dp [MAX]; int c [MAX], w [MAX]; int v; void ZeroOnePack (int cost, int wei) // 01 {int I; for (I = v; I> = cost; I --) {dp [I] = max (dp [I], dp [I-cost] + wei) ;}} void CompletePack (int cost, int wei) // completely {int I; for (I = cost; I <= v; I ++) {dp [I] = max (dp [I], dp [I-cost] + wei) ;}} void MultiplePack (int cost, int wei, int cnt) // multiple {If (v <= cnt * cost) // if the total capacity is smaller than the total capacity of the item, the item can be retrieved until it is finished, equivalent to a full backpack {CompletePack (cost, wei ); return;} else // otherwise multiple backpacks will be converted to 01 backpack {int k = 1; while (k <= cnt) {ZeroOnePack (k * cost, k * wei ); cnt = cnt-k; k = 2 * k;} ZeroOnePack (cnt * cost, cnt * wei) ;}} int main () {int n; while (~ Scanf ("% d", & n, & v), n + v) {int I; for (I = 0; I <n; I ++) scanf ("% d", & c [I]); for (I = 0; I <n; I ++) scanf ("% d ", & w [I]); memset (dp, 0, sizeof (dp); for (I = 0; I <n; I ++) {MultiplePack (c [I], c [I], w [I]);} int sum = 0; for (I = 1; I <= v; I ++) {if (dp [I] = I) {sum ++ ;}} printf ("% d \ n", sum) ;}return 0 ;}