HDU 2844 coins (motion gauge)

Coins Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 6904 accepted submission (s): 2809


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
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This is a good question! At the beginning, writing with the original multiple backpacks always times out. Later I heard that binary optimization can be used. I just copied the information to get rid of it !~
Question:
Give you a silver coin's nominal value and corresponding number, then give a number M and ask how many coins from 1 to m can be expressed with existing coins.

A simple multi-bag, using binary to optimize the face value of a coin. Then, determine the number of packages that meet B [I] = I;

A [I] indicates the value of the coin, and B [I] indicates the number of coins.

# Include <iostream> # include <algorithm>Using namespaceSTD;# Define M 100005IntDP[M],Vis[M]; // If you want to use a 01 backpack, the system will time out... IntA[105],B[105]; IntN,TOT,M; IntMax(IntX, IntY) {ReturnX>Y?X:Y;} // Baidu income, worship. VoidCompletepack(IntCost, IntWeight) // If you cannot finish it, it is a complete backpack. {For (intI=Cost;I<=M;I++)DP[I] =Max(DP[I],DP[I-Cost] +Weight);} VoidZeroonepack(IntCost, IntWeight) // If it can be obtained, it is the 01 backpack. {For (intI=M;I> =Cost;I--)DP[I] =Max(DP[I],DP[I-Cost] +Weight);} VoidMultiplypack(IntCost, IntWeight, IntAmount) // Multiple backpacks. {If (Cost*Amount> =M)// Not complete. Completepack(Cost,Weight); Else {intK=1; While (K<Amount){Zeroonepack(K*Cost,K*Weight);// The legendary binary optimization, indicating that it is not clear. I think it is a time to take 1, 2, 4, 8... Coin, multiply thoughts. Amount-=K;K<=1;}Zeroonepack(Amount*Cost,Amount*Weight); // Calculate the remaining items here. } Int main (intI, IntJ, IntK) {While (Scanf("% D",&N,&M)! =EOF&&N&&M){TOT=K=0; (I=1;I<=N;I++)Scanf("% D",&A[I]); (I=1;I<=N;I++)Scanf("% D",&B[I]);Memset(DP,0, Sizeof (DP); (I=1;I<=N;I++)Multiplypack(A[I],A[I],B[I]); (I=1;I<=M;I++) If (DP[I] =I)TOT++;Printf("% D \ n",TOT);} Return0;}