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  • Codeforces 189A:Cut Ribbon(完全背包,DP)

    time limit per test : 1 second

    memory limit per test : 256 megabytes

    input : standard input

    output : standard output

    Polycarpus has a ribbon, its length is nn. He wants to cut the ribbon in a way that fulfils the following two conditions:

    • After the cutting each ribbon piece should have length aa, bb or cc.
    • After the cutting the number of ribbon pieces should be maximum.

    Help Polycarpus and find the number of ribbon pieces after the required cutting.

    Input

    The first line contains four space-separated integers nn, aa, bb and cc (1n,a,b,c4000)(1 ≤ n, a,b, c≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers aa, bb and cc can coincide.

    Output

    Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.

    Examples

    input

    5 5 3 2
    

    output

    2
    

    input

    7 5 5 2
    

    output

    2
    

    Note

    In the first example Polycarpus can cut the ribbon in such way: the first piece has length 22, the second piece has length 33.

    In the second example Polycarpus can cut the ribbon in such way: the first piece has length 55, the second piece has length 22.

    Solve

    完全背包,看成给出三种商品,恰好能够装满背包的情况

    注意初始化的问题,如果把dpdp数组全部初始化为1-1的话,需要注意dp[j]=1&&dp[ja[i]]=1dp[j]=-1&&dp[j-a[i]]=-1的情况。或者就全部初始值给成小于1-1的数

    Code

    /*************************************************************************
    
    	 > Author: WZY
    	 > School: HPU
    	 > Created Time:   2019-04-01 18:30:38
    	 
    ************************************************************************/
    #include <cmath>
    #include <cstdio>
    #include <time.h>
    #include <cstring>
    #include <limits.h>
    #include <iostream>
    #include <algorithm>
    #include <random>
    #include <iomanip>
    #include <map>
    #include <set>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <string>
    #include <random>
    #define ll long long
    #define ull unsigned long long
    #define lson o<<1
    #define rson o<<1|1
    #define ms(a,b) memset(a,b,sizeof(a))
    #define SE(N) setprecision(N)
    #define PSE(N) fixed<<setprecision(N)
    #define bug cerr<<"-------------"<<endl
    #define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"
    "
    #define LEN(A) strlen(A)
    const double E=exp(1);
    const double eps=1e-9;
    const double pi=acos(-1.0);
    const int mod=1e9+7;
    const int maxn=1e6+10;
    const int maxm=1e3+10;
    const int moha=19260817;
    const int inf=1<<30;
    const ll INF=1LL<<60;
    using namespace std;
    inline void Debug(){cerr<<'
    ';}
    inline void MIN(int &x,int y) {if(y<x) x=y;}
    inline void MAX(int &x,int y) {if(y>x) x=y;}
    inline void MIN(ll &x,ll y) {if(y<x) x=y;}
    inline void MAX(ll &x,ll y) {if(y>x) x=y;}
    template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
    	cerr<<arg<<"";Debug(rest...);}
    int dp[maxn];
    int main(int argc, char const *argv[])
    {
    	ios::sync_with_stdio(false);cin.tie(0);
    	cout.precision(20);
    	#ifndef ONLINE_JUDGE
    	    freopen("in.txt", "r", stdin);
    	    freopen("out.txt", "w", stdout);
    		srand((unsigned int)time(NULL));
    	#endif
    	int n;
    	int a[4];
    	cin>>n>>a[0]>>a[1]>>a[2];
    	ms(dp,-1);
    	dp[0]=0;
    	for(int i=0;i<3;i++)
    		for(int j=a[i];j<=n;j++)
    			if(dp[j-a[i]]!=-1)
    				MAX(dp[j],dp[j-a[i]]+1);
    	cout<<dp[n]<<endl;
    	#ifndef ONLINE_JUDGE
    	    cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s.
    ";
    	#endif
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Friends-A/p/11054959.html
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