Luo Gu P2725 Stamps Stamps

https://www.luogu.org/problem/P2725

JDOJ 1797: Stamps Stamps

https://neooj.com:8082/oldoj/problem.php?id=1797

Description

Given a set of N denominations of stamps (e.g., {1, 3,}), and an upper limit K - K represents a stamp can be affixed on the envelope calculated maximum postage continuously posted from 1 to M.
E.g. , assuming there are 3 points and 1 point stamps; you can stick easily up to five posted stamps 1 to 5 minutes postage (1 minute by the stamps on the line), the next postage is not difficult:
6 = 3 + 3
. 7 = 3 + 3 + 1
8 = 3 + 3 + 1 + 1
9 = 3 + 3 + 3
10 = 3 + 3 + 3 + 1
11 = 3 + 3 + 3 + 1 + 1
12 = 3 + 3 3 + 3 +
13 = 3 + 3 + 3 + 3 + 1.
However, the use of 5 or 1 min 3 min 14 min stamp impossible to put postage. Thus, for both the stamps and the set upper limit K = 5, the answer is M = 13.

Input

Line 1: two integers, K and NK (1 <= K <= 200) is the total number of available stamps .N (1 <= N <= 50) is the number of postage stamp.
End of the line 2 .. File: N integers, each row 15, lists all of the N denomination stamps, face value not more than 10,000.

Output

Line 1: an integer from 1 pm start contiguous set of available no more than the number of postage stamps K posted.

Sample Input

5 2 1 3

Sample Output

Complete knapsack problem, I personally feel that this question is that the pits did not give a specific date range:

I assume dp [i] represents the number of postage stamp sheets when i needed.

Then trouble, postage how to do?

At first I put an upper limit is set to 10000001 results have a point TLE, one-tenth of that I shrink, resulting in two points WA, then I really did not move put up in accordance with the guidance of a solution to a problem one-fifth, AC up.

JDOJ then hung up.

This is who hold up ah! !

Attach two codes, first Los Valley AC, second is JDOJ AC, small partners for reference.

#include <cstdio> 
#include <algorithm>
 the using  namespace STD;
 int K, n-;
 int DP [ 2000001 ]; // DP [i] represents the number of denominations of stamps when the minimum needed i 
int main () 
{ 
    Scanf ( " % D% D " , & K, & n-);
     for ( int I = . 1 ; I <= 2000001 ; I ++ ) 
        DP [I] = 2147483647 ; 
    DP [ 0 ] = 0 ;
     for ( int I = . 1 ; I <= n; i ++)
    {
        int a;
        scanf("%d",&a);
        for(int j=a;j<=2000001;j++)
            if(dp[j-a]+1<=k)
                dp[j]=min(dp[j],dp[j-a]+1);
    }
    for(int i=1;i<=2000001;i++)
        if(dp[i]==2147483647)
        {
            printf("%d",i-1);
            return 0;
        }
}

JDOJ

#include<iostream>
#include<algorithm>
#include<cstring>
#define maxv 2000005
#define inf 0x3f3f3f3f
using namespace std;
int n,k,ans,mx,c[55];
int dp[maxv];
int main()
{
    cin>>k>>n;
    memset(dp,inf,sizeof(dp));
    dp[0]=0;
    for(int i=1;i<=n;i++) 
        cin>>c[i];
    sort(c+1,c+n+1);
    for(int i=1;i<=n;i++)
    {
        mx=c[i]*k;
        for(int j=c[i];j<=mx;j++)
            if(dp[j-c[i]]<min(k,dp[j]-1))
                dp[j]=dp[j-c[i]]+1;
    }
    for(int i=1;i<maxv;i++)
    {
        if(dp[i]!=inf) 
            ans++;
        else 
            break;
    } 
    Cout << years;
    return  0 ; 
}

USACO Stamps