题意描述
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.
思路
很明显的一道多重背包的题目，由于数据范围过大，所以不能使用朴素解法，可以使用二进制优化来解决。设dp[i]表示金币能否拼成i块钱。最后遍历一遍数组即可。
AC代码
代码语言：javascript复制#include<iostream>
#include<string>
#include<cstring>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x )
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=150100;
const int INF=0x3f3f3f3f;
int v[N],dp[N],w[N];
void solve(){
int n,m;
while(cin>>n>>m && n && m){
rep(i,0,m 1) dp[i]=0;
rep(i,1,n 1) cin>>v[i];
int cnt=1;
rep(i,1,n 1){
int c;cin>>c;
int k=1;
while(c>=k){
w[cnt ]=v[i]*k;
c-=k;
k*=2;
}
if(c>0){
w[cnt ]=v[i]*c;
}
}
n=cnt;
dp[0]=1;
rep(i,1,n){
rrep(j,m,w[i]){
dp[j]=max(dp[j],dp[j-w[i]]);
}
}
int ans=0;
rep(i,1,m 1) if(dp[i]>0) ans ;
cout<<ans<<endl;
}
}
int main(){
IOS;
solve();
return 0;
}
