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 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.
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 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0
看到英问题就头疼！！！
这个题目看懂题目，理解题意是我觉得最难的一步！（英语贼烂）
其实这个题目意思就是在这个[1,m]这个区间里面有多少个背包可以装满。
（题意浓缩起来只有这么点，提炼出来真心费力）
这个题目第二难的一部就是进行二进制优化。
（由于本菜鸟刚学DP确实这个对我而言是一个长期理解的过程）
二进制优化：
任何一个数都可以由2的N次方来表示，例如4可以有1，2来表示；
8可以有1，2，4来表示；11可以有1，2，4，8来表示。
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 #define maxn 100010 7 int dp[maxn],a[105],b[105]; 8 int main() 9 { 10 int n,m; 11 while(scanf("%d%d",&n,&m)!=EOF){ 12 if (n==0 && m==0) break; 13 memset(dp,0,sizeof(dp)); 14 for (int i=1 ;i<=n ;i++) 15 scanf("%d",&a[i]); 16 for (int i=1 ;i<=n ;i++) 17 scanf("%d",&b[i]); 18 for (int i=1 ;i<=n ;i++){ 19 if (a[i]*b[i]>m) { 20 for (int k=0;k<=m ;k++){ 21 if (k>=a[i]) dp[k]=max(dp[k],dp[k-a[i]]+a[i]); 22 } 23 }else { 24 for (int j=1 ;j<=b[i] ;j=j*2){ 25 for (int k=m ; k>=a[i]*j ;k--){ 26 dp[k]=max(dp[k],dp[k-a[i]*j]+a[i]*j); 27 } 28 b[i]-=j; 29 } 30 if (b[i]>0) { 31 for (int k=m ;k>=a[i]*b[i] ;k--) { 32 dp[k]=max(dp[k],dp[k-a[i]*b[i]]+a[i]*b[i]); 33 } 34 } 35 } 36 } 37 int sum=0; 38 for (int i=1 ;i<=m ;i++){ 39 if (dp[i]==i) sum++; 40 } 41 printf("%d\n",sum); 42 } 43 return 0; 44 }
