• 3.1.6 Stamps


    Stamps

    Given a set of N stamp values (e.g., {1 cent, 3 cents}) and an upper limit K to the number of stamps that can fit on an envelope, calculate the largest unbroken list of postages from 1 cent to M cents that can be created.

    For example, consider stamps whose values are limited to 1 cent and 3 cents; you can use at most 5 stamps. It's easy to see how to assemble postage of 1 through 5 cents (just use that many 1 cent stamps), and successive values aren't much harder:

    • 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, there is no way to make 14 cents of postage with 5 or fewer stamps of value 1 and 3 cents. Thus, for this set of two stamp values and a limit of K=5, the answer is M=13.

    The most difficult test case for this problem has a time limit of 3 seconds.

    PROGRAM NAME: stamps

    INPUT FORMAT

    Line 1: Two integers K and N. K (1 <= K <= 200) is the total number of stamps that can be used. N (1 <= N <= 50) is the number of stamp values.
    Lines 2..end: N integers, 15 per line, listing all of the N stamp values, each of which will be at most 10000.

    SAMPLE INPUT (file stamps.in)

    5 2
    1 3
    

    OUTPUT FORMAT

    Line 1: One integer, the number of contiguous postage values starting at 1 cent that can be formed using no more than K stamps from the set.

    SAMPLE OUTPUT (file stamps.out)

    13

    /*
    ID:makeeca1
    PROG:stamps
    LANG:C++
    */
    #include<cstdio>
    #define min(x,y) (x>y?y:x)
    using namespace std;
    #define MAX 2100001
    int f[MAX],a[100],n,m;
    int main(){
        freopen("stamps.in","r",stdin);
        freopen("stamps.out","w",stdout);
        scanf("%d%d",&m,&n);
        for (int i=1;i<=n;i++)scanf("%d",&a[i]);
        f[0]=0;
        for (int i=0;;i++){
            if ((f[i]||i==0)&&f[i]<=m){
                for (int j=1;j<=n;j++)
                    if (f[i+a[j]])
                        f[i+a[j]]=min(f[i+a[j]],f[i]+1);
                    else f[i+a[j]]=f[i]+1;
            }else {printf("%d
    ",i-1);return 0;}
        }
    }
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  • 原文地址:https://www.cnblogs.com/makeecat/p/3289359.html
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