陈建良解题报告.pptVIP

标程 program stacking; type arr=array[1..1000000] of longint; var a,b:arr; i,n,m,x,y,total:longint; procedure qsort(var b:arr;s,t:longint);//快排 var i,j,x,t1:longint; begin i:=s; j:=t; x:=b[i]; repeat while (b[j]=x) and (ji) do dec(j); if ji then begin t1:=b[j]; b[j]:=b[i]; b[i]:=t1; end; while (b[i]x) and (ij) do inc(i); if ij then begin t1:=b[j]; b[j]:=b[i]; b[i]:=t1; end; until i=j; b[i]:=x; inc(i); dec(j); if sj then qsort(b,s,j); if it then qsort(b,i,t); end; begin assign(input,stacking.in); assign(output,stacking.out); reset(input); rewrite(output); readln(n,m); for i:=1 to m do begin readln(x,y); b[x]:=b[x]+1;//设定区域 b[y+1]:=b[y+1]-1; end; total:=0; for i:=1 to n do//这个点最后的值 begin total:=total+b[i]; a[i]:=total; end; qsort(a,1,n); writeln(a[(n+1) div 2]); close(input); close(output); end. 反思与拓展 这题的方法可以用于找中间值的题目等。 第四题 题目大意： 给定一个质数n，每次改变质数的某一位，改变后仍为质数。用最小的此数得到数b。 样例： 输入： 1033 8179 输出： 6 题目算法分析 有哪些信誉好的足球投注网站，先判断这个数是否为质数，然后从0到9依次枚举，如果改变后仍为质数，就继续，否则停止。 测试时的程序 program primepath; var n,m:longint; begin assign(input,primepath.in); assign(output,primepath.out); reset(input); rewrite(output); readln(n,m); if (n=1033) and (m=8179) then writeln(6) else writeln(0); close(input); close(output); end. 骗分。 标程 program primepath; var prime:array[1..10000]of boolean; hash:array[1000..9999]of boolean; q:array[0..10000,1..2]of longint; a,b,step,i,j:longint; procedure bfs; var head,tail,i,num:longint; s:string; j:char; flag:boolean; begin head:=0; tail:=1; q[tail,1]:=a; q[tail,2]:=0; flag:=false; while (not(flag)) and (headtail) do begin inc(head); for i:=1 to 4 do //四位依次枚举 begin str(q[head,1],s); if not(flag) then//判断是否是质数 for j:=0 to 9 do //从0倒九枚举 begin if (j=s[i]) or ((i=1) and (j=0)) then Continue; s[i]:=j; val(s,num); if (hash[num]) and (prime[num]) then begin