• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            USACO Section 3.1 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
                
            Analysis

            This problem is also a DP problem. Considering the minimum volumn of a bag that cannot be stored by the K stamps and N kinds of the given stamps, we can find the problem becomes to find the dynamic function to describe it. We call f[i] as the minimum number of stamps to get the value of i, which depends on the minimum of a set of f[i-value[j]]+1),((i-Value[j]>=0 j=1..Stamps).

            In order to abbreviate functioning amount, find out the MaxValue, which is the maximum of set Value[i], and all of its multiples can be only fixed with i/MaxValue.

            At last,the edge of the problem is when f[k]>maxuse and print k-1.

            Code

            /*
            ID:braytay1
            PROG:stamps
            LANG:C++
            */

            #include 
            <iostream>
            #include 
            <fstream>
            #define min(a,b) (a<b)?a:b
            using namespace std;


            int main(){
                ifstream fin(
            "stamps.in");
                ofstream fout(
            "stamps.out");
                
            int k1,n,max=-1;
                
            int s[51];
                
            short int f[2000000];
                fin
            >>k1>>n;
                
            for (int i=1;i<=n;i++{fin>>s[i];max=(max>s[i])?max:s[i];}
                
            if (max==1{fout<<max*k1<<endl;return 0;}
                memset(f,
            100,sizeof(f));
                f[
            0]=0;
                
            for (int j=1,k=max;k<=max*k1;k+=max,j++) f[k]=j;
                
            int kk;
                
            for (kk=1;kk<=max*k1;++kk){
                    
            if (kk%max==0continue;
                    
            for (int j=1;j<=n;++j) {
                        
            if (kk-s[j]>=0
                            
            if (f[kk-s[j]]+1<f[kk]) f[kk]=f[kk-s[j]]+1;
                    }

                    
            if (f[kk]>k1) break;
                }

                fout
            <<kk-1<<endl;
                
            return 0;
            }


             

            posted on 2008-08-22 01:32 幻浪天空領主 閱讀(220) 評論(0)  編輯 收藏 引用 所屬分類: USACO

            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            導航

            統計

            常用鏈接

            留言簿(1)

            隨筆檔案(2)

            文章分類(23)

            文章檔案(22)

            搜索

            最新評論

            閱讀排行榜

            評論排行榜

            精品国产VA久久久久久久冰| 色诱久久av| 久久久久国产一区二区三区| 欧美精品丝袜久久久中文字幕| 无码人妻久久一区二区三区蜜桃| 亚洲精品国精品久久99热一| 色综合久久最新中文字幕| 亚洲乱码日产精品a级毛片久久| 久久综合噜噜激激的五月天| 国内精品久久久久久中文字幕 | 很黄很污的网站久久mimi色| 国产aⅴ激情无码久久| 久久99精品久久久久久不卡| 色偷偷久久一区二区三区| 久久综合偷偷噜噜噜色| 国产精品gz久久久| 久久精品www| 97久久精品无码一区二区天美| 九九精品久久久久久噜噜| 久久成人18免费网站| 香蕉久久一区二区不卡无毒影院| 婷婷久久久亚洲欧洲日产国码AV | 久久精品一区二区国产| 久久午夜福利无码1000合集| 久久久噜噜噜久久| 国内精品久久久久久久涩爱| 久久九九有精品国产23百花影院| 久久精品天天中文字幕人妻| 无遮挡粉嫩小泬久久久久久久| 伊人久久大香线蕉综合网站| 久久亚洲欧洲国产综合| 国产午夜精品久久久久九九| 欧美久久综合性欧美| 99久久精品无码一区二区毛片| 久久精品国产亚洲综合色| 99久久精品国内| 国产91久久综合| 久久中文精品无码中文字幕| 一级女性全黄久久生活片免费 | 夜夜亚洲天天久久| 久久精品国产一区二区三区|