• <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 4.1 Beef McNuggets

            這題有些難。雖然知道是動態規劃題,但是不知道要開多大的數組,后來看analysis用一個256大小的數組循環使用,方法很巧妙。
            先將box進行排序。
            如果box里面的數的最大公約數不為1的話,那么所有組成的數,只可能是這個公約數的倍數,因此沒有上限,輸出為0.
            用last記錄最小的“不能組成的數”。這樣當last之后有boxs[0]個連續數都可以組成的話,那么所有的數都可以組成。
            last+1...last+box[0]可以組成的話,那么每個數都加一個box[0],那么新一輪的box[0]個數也可以組成,以此類推。

            #include?<iostream>
            #include?
            <fstream>

            using?namespace?std;

            ifstream?fin(
            "nuggets.in");
            ofstream?fout(
            "nuggets.out");

            #ifdef?_DEBUG
            #define?out?cout
            #define?in?cin
            #else
            #define?out?fout
            #define?in?fin
            #endif

            int?box_num;
            int?boxs[10];

            bool?ok[256];

            int?gcd(int?a,int?b)
            {
            ????
            if(a<b)?swap(a,b);

            ????
            int?tmp;

            ????
            while(b!=0){
            ????????tmp?
            =?a;
            ????????a?
            =?b;
            ????????b?
            =?tmp%b;
            ????}

            ????
            return?a;
            }

            void?solve()
            {

            ????
            in>>box_num;
            ????
            for(int?i=0;i<box_num;++i)
            ????????
            in>>boxs[i];

            ????sort(
            &boxs[0],&boxs[box_num]);
            ????
            ????
            int?t?=?boxs[0];

            ????
            for(int?i=1;i<box_num;++i){
            ????????t?
            =?gcd(t,boxs[i]);
            ????}

            ????
            if(t!=1){
            ????????
            out<<0<<endl;
            ????????
            return;
            ????}

            ????memset(ok,
            0,sizeof(ok));

            ????
            int?last?=?0;
            ????ok[
            0]?=?true;
            ????
            int?i=0;

            ????
            while(true){
            ????????
            if(ok[i%256]){
            ????????????ok[i
            %256]?=?0;
            ????????????
            if(i-last>=boxs[0]){
            ????????????????
            out<<last<<endl;
            ????????????????
            return;
            ????????????}
            ????????????
            for(int?x=0;x<box_num;++x){
            ????????????????ok[(i
            +boxs[x])%256]?=?true;
            ????????????}
            ????????}
            else{
            ????????????last?
            =?i;
            ????????}
            ????????
            ++i;
            ????}
            }

            int?main(int?argc,char?*argv[])
            {
            ????solve();?
            ????
            return?0;
            }


            Beef McNuggets

            Hubert Chen

            Farmer Brown's cows are up in arms, having heard that McDonalds is considering the introduction of a new product: Beef McNuggets. The cows are trying to find any possible way to put such a product in a negative light.

            One strategy the cows are pursuing is that of `inferior packaging'. ``Look,'' say the cows, ``if you have Beef McNuggets in boxes of 3, 6, and 10, you can not satisfy a customer who wants 1, 2, 4, 5, 7, 8, 11, 14, or 17 McNuggets. Bad packaging: bad product.''

            Help the cows. Given N (the number of packaging options, 1 <= N <= 10), and a set of N positive integers (1 <= i <= 256) that represent the number of nuggets in the various packages, output the largest number of nuggets that can not be purchased by buying nuggets in the given sizes. Print 0 if all possible purchases can be made or if there is no bound to the largest number.

            The largest impossible number (if it exists) will be no larger than 2,000,000,000.

            PROGRAM NAME: nuggets

            INPUT FORMAT

            Line 1: N, the number of packaging options
            Line 2..N+1: The number of nuggets in one kind of box

            SAMPLE INPUT (file nuggets.in)

            3
            3
            6
            10

            OUTPUT FORMAT

            The output file should contain a single line containing a single integer that represents the largest number of nuggets that can not be represented or 0 if all possible purchases can be made or if there is no bound to the largest number.

            SAMPLE OUTPUT (file nuggets.out)

            17

            posted on 2009-07-12 14:58 YZY 閱讀(641) 評論(0)  編輯 收藏 引用 所屬分類: AlgorithmUSACO動態規劃

            導航

            <2009年7月>
            2829301234
            567891011
            12131415161718
            19202122232425
            2627282930311
            2345678

            統計

            常用鏈接

            留言簿(2)

            隨筆分類

            隨筆檔案

            搜索

            積分與排名

            最新評論

            閱讀排行榜

            久久免费99精品国产自在现线| 日韩AV无码久久一区二区| 青青热久久国产久精品 | 久久高清一级毛片| 伊人久久大香线蕉无码麻豆| 亚洲精品无码久久久久久| 久久99精品久久久久久久不卡| 久久天堂AV综合合色蜜桃网| 久久久久久免费一区二区三区| 国产精品综合久久第一页| 日本WV一本一道久久香蕉| 午夜精品久久久久久99热| 久久国产精品免费一区二区三区| 日本久久久久久久久久| 久久久久久亚洲AV无码专区| 亚洲乱亚洲乱淫久久| 国产成人精品综合久久久久| 久久精品国产福利国产秒| 久久精品综合网| 99精品伊人久久久大香线蕉| 2021最新久久久视精品爱| 欧美日韩中文字幕久久伊人| 亚洲国产成人精品久久久国产成人一区二区三区综 | 久久久噜噜噜久久中文字幕色伊伊| 亚洲国产精品久久久天堂| 国内精品久久久久久久coent | 狠狠色丁香婷婷久久综合不卡| 午夜视频久久久久一区 | 久久精品国产AV一区二区三区| 青青青青久久精品国产| 久久av无码专区亚洲av桃花岛| 伊人久久大香线蕉综合网站| 婷婷综合久久中文字幕| av无码久久久久久不卡网站| 久久人人爽人人爽人人片av麻烦| 久久99精品免费一区二区 | 亚洲国产成人精品女人久久久 | 午夜天堂精品久久久久| 久久久久久精品免费看SSS| 午夜精品久久久久久久无码| 久久国产精品一区|