POJ 1742 coins（背包+二进制优化+bitset）

最新推荐文章于 2019-10-28 23:54:22 发布

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本文介绍了一道关于硬币组合的编程题目，通过使用bitset和二进制优化的方法解决了一个典型的多重背包问题。目的是计算出特定数量和类型的硬币能够组成的从1到m之间的不同价格数目。

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coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 39107		Accepted: 13266

Description

People in Silverland use coins.They have coins of value A1,A2,A3…An Silverland dollar.One day Tony opened his money-box 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 10 

1 2 4 2 1 1 

2 5 

1 4 2 1 

0 0

Sample Output

8 

4

Source

LouTiancheng@POJ

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题意

给你n,m，表示有n种硬币，接下来2n个数，前n个数表示n个硬币的价格，后n个数表示每种硬币各有几个。求这n种硬币能组成1~m中多少个价格。

思路

这题显然是个多重背包，但是很显然，直接多重背包肯定会T，所以我们可以用bitset+二进制优化来优化它。二进制优化就是将硬币的个数cnt分成 $cnt=2^0+2^1+2^2...+rest$ ，rest就是剩下的不满二的次方的那部分。因为这样可以将1~cnt的所有方法给组出来。还可以加快速度。对于目前的price，显然如果价格x可以被组出来的话，当且仅当x-price可以被组出来。所以我们可以用bitset的÷32的特性来进行常数优化。

Code

#pragma GCC optimize(3)
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cctype>
#include<climits>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<climits>
#include<vector>
#include<bitset>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
bitset<100010> get;
int n,m,a[105],c;
inline void solve(int x){get|=get<<x;}
int main() {
    while(1) {
        get.reset();
        readInt(n);
        readInt(m);
        if(!n&&!m)
            break;
        get[0]=1;
        for(int i=1;i<=n;i++)
            readInt(a[i]);
        for(int i=1;i<=n;i++) {
            readInt(c);
            for(int p=1;p<c;p<<=1) {
                solve(a[i]*p);
                c-=p;
            }
            solve(a[i]*c);
        }
        int ans=0;
        for(int i=1;i<=m;i++)
            if(get[i])
                ans++;
        printf("%d\n",ans);
    }
    return 0;
}