CoinsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8052Accepted Submission(s): 3291
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 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
3 101 2 4 2 1 12 51 4 2 10 0
Sample Output
84
最近一直在做DP的题目，，都是一些模板题，，现在已成了背包九讲作者的死忠了。如果我是女的，，，我一定要嫁给写出背包九讲的男人o(╯□╰)o
先说一下二进制优化的原理：
1、2、4可以组合出所有小于8的数；1、2、4、8可以组合出所有小于16的数；1、2、4、8、16可以组合出所有小于32的数；
我的代码：
#include <stdio.h>#include <string.h>#define M 100100 #define N 110int a[N] , c[N] , dp[M] ;void zeroOnePack(int value , int m){for(int i = m ; i >= value ; –i){if(dp[i]<dp[i-value]+value){dp[i] = dp[i-value]+value ;}}}void completePack(int value , int m){for(int i = value ; i <= m ; ++i){if(dp[i]<dp[i-value]+value){dp[i] = dp[i-value]+value ;}}}void multiplePack(int value , int count , int total){if(value*count>total){completePack(value , total) ;}else{int k = 1 ;while(k<=count){zeroOnePack(k*value , total) ;count -= k ;k = 2*k ;}zeroOnePack(count*value,total) ;}}int main(){int n ,m ;while(~scanf("%d%d",&n,&m) && (n||m)){for(int i = 0 ; i < n ; ++i){scanf("%d",&a[i]);}for(int i = 0 ; i < n ; ++i){scanf("%d",&c[i]);}memset(dp,0,sizeof(dp)) ;for(int i = 0 ; i < n ; ++i){multiplePack(a[i],c[i],m);}int ans = 0 ;for(int i = 1 ; i <= m ; ++i){if(i == dp[i]){++ans ;}}printf("%d\n",ans) ;}return 0 ;}
即使爬到最高的山上，一次也只能脚踏实地地迈一步。
