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9-25 解题报告

时间:2015-09-25 21:42:08      阅读:221      评论:0      收藏:0      [点我收藏+]

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1.catfish

var
        n,k,ans,tot:int64;
        i,j:longint;
        a:array[0..100] of int64;
        f:array[0..40] of int64;

        begin
                assign(input,catfish.in);
                reset(input);
                assign(output,catfish.out);
                rewrite(output);
                readln(n);        f[1]:=1;
                for i:=2 to 31 do f[i]:=3*f[i-1];
                for j:=1 to n do
                begin
                        read(k); tot:=0;
                        ans:=0;
                        while k<>0 do
                        begin
                                inc(tot);
                                a[tot]:=k mod 2;
                                k:=k div 2;
                        end;
                        for i:=tot downto 1 do
                        if a[i]=1 then inc(ans,f[i]);
                        writeln(ans);
                end;
                close(input);
                close(output);
        end.

2.meal

var
        map,f,up,left,right,leftup,rightup,f1,f2:array[0..1,0..2505] of longint;
        temp,i,j,ii,k,l,m,n,max,t:longint;


        function min(a,b,c:longint):longint;
        begin
                if a<b then min:=a
                else min:=b;
                if c<min then min:=c;
        end;

        begin
                assign(input,meal.in);
                reset(input);
                assign(output,meal.out);
                rewrite(output);
                while not eof do
                begin
                        fillchar(up,sizeof(up),0);
                        fillchar(left,sizeof(up),0);
                        fillchar(right,sizeof(up),0);
                        fillchar(f1,sizeof(up),0);
                        fillchar(f2,sizeof(up),0);
                        fillchar(map,sizeof(up),10);

                        readln(n,m);
                        max:=0;
                        t:=1;
                        for i:=1 to m do
                        begin
                                read(map[t][i]);
                                if map[t][i]=1 then
                                begin
                                        rightup[t][i]:=1;
                                        leftup[t][i]:=1;
                                        f1[t][i]:=1;
                                        f2[t][i]:=1;
                                        max:=1;
                                        continue;
                                end;
                                up[t][i]:=1;
                        end;
                        readln;
                        for ii:=2 to n do
                        begin
                                t:=t xor 1;
                                for i:=1 to m do
                                begin
                                        left[t][i]:=0;
                                        right[t][i]:=0;
                                        up[t][i]:=0;
                                        leftup[t][i]:=0;
                                        rightup[t][i]:=0;
                                        f1[t][i]:=0;
                                        f2[t][i]:=0;
                                end;

                        for i:=1 to m do
                        begin
                                read(map[t][i]);
                                if map[t][i]=1 then
                                begin
                                        f1[t][i]:=1;
                                        f2[t][i]:=1;
                                        continue;
                                end;
                                left[t][i]:=1;
                                right[t][i]:=1;
                                up[t][i]:=1;
                        end;
                        readln;

                        for i:=1 to m do if map[t][i]=0 then
                        begin
                                if map[t][i-1]=0 then left[t][i]:=left[t][i-1]+1;
                                if map[t xor 1,i]=0 then up[t][i]:=up[t xor 1,i]+1;
                        end;

                        for i:=m downto 1 do if map[t][i]=0 then
                        begin
                                if map[t][i+1]=0 then right[t][i]:=right[t][i+1]+1;
                        end;

                        for i:=1 to m do if (map[t][i]=1) and (map[t xor 1][i-1]=1) then
                        begin
                                f1[t][i]:=min(left[t][i-1],up[t xor 1][i],f1[t xor 1][i-1])+1;
                                if max<f1[t][i] then max:=f1[t][i];
                        end;

                        for i:=1 to m do if (map[t][i]=1) and (map[t xor 1][i+1]=1) then
                        begin
                                f2[t][i]:=min(right[t][i+1],up[t xor 1][i],f2[t xor 1,i+1])+1;
                                if max<f2[t][i] then max:=f2[t][i];
                        end;
                        end;
                        writeln(max);
                end;
                close(input);
                close(output);
        end.

第三题指针,真心不会,熟悉pascal的大神求包养QvO

 喜欢就收藏一下,vic私人qq:1064864324,加我一起讨论问题,一起进步^-^

9-25 解题报告

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原文地址:http://www.cnblogs.com/victorslave/p/4839517.html

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