NOIP2012普及组复赛解题报告

1、第一题：筛选法求素数。var a:array1.50000of boolean; x,y,i,j,n:longint;beginassign(input,'prime.in'); reset(input);assign(output,'prime.ans'); rewrite(output);readln(n);fillchar(a,sizeof(a),true);for i:=2 to trunc(sqrt(50000) do if ai then for j:=2 to 50000 div i do ai*j:=false;for i:=2 to 50000

2、 do if ai and (n mod i=0) then begin writeln(n div i); break; end;close(input); close(output);end.第二题：模拟法，因为指示牌上的数字较大，可先统计每层有楼梯的个数，再取余。var a,b:array0.10000,0.100of longint; x,y,i,j,k,n,m,s,t,w:longint; c:array1.10000of integer;begin assign(input,'c:1.in'); reset(input); assign(input,'tre

3、asure.in'); reset(input); assign(output,'treasure.ans'); rewrite(output); readln(n,m); for i:=1 to n do begin s:=0; for j:=0 to m-1 do begin readln(ai,j,bi,j); if ai,j=1 then s:=s+1; end; ci:=s; end; readln(x); s:=0; for i:=1 to n do begin y:=x; s:=(s+bi,x)mod 20123; t:=bi,xmod ci+ci; re

4、peat if ai,y=1 then t:=t-1; if t=0 then break; y:=y+1; if y=m then y:=0; until false; x:=y; end; writeln(s); close(input); close(output);end.第三题：动态规划FI，J 表示 用I种花摆J盆共有和摆法，则FI，J=（ FI-1，J ），（注意只取J-I的盆数J盆求和）var a:array0.100,0.100of longint; i,j,k,m,n,s,t:longint; w:array0.100of longint;begin assign(inpu

5、t,'flower.in'); reset(input); assign(output,'flower.ans'); rewrite(output); readln(n,m); for i:=1 to n do read(wi); for i:=1 to w1 do a1,i:=1; for i:=0 to m do ai,0:=1; for i:=2 to m do begin for j:=1 to m do begin s:=0; for k:=j-wi to j do if k>=0 then s:=(s+ai-1,k)mod 1000007; a

6、i,j:=s; end; end; writeln(am,m); close(input); close(output);end.4、搜索，可是第10个数据过不了，要输出“-1”，不知道哪里BUG了，不想了。var w:array1.100of longint; a,b:array1.100,1.100of longint; i,j,k,m,n,p,q,r,s,t,sum,min:longint; f,v:array1.100of boolean;procedure try(x:integer);var i,j,k:integer; flag:boolean;begin if x=t then

7、 begin if (sum<>0)and(sum<min) then min:=sum; end else begin for i:=1 to n do begin if fi and (bx,i<>0) then begin flag:=true; for j:=1 to n do if (vj=true) or (ai,j=1) then begin flag:=false; break; end; fi:=false; vwi:=true; sum:=sum+bx,i; if sum<min then try(i); fi:=true; vwi:=f

8、alse; sum:=sum-bx,i; end; end; end;end;begin assign(input,'culture.in'); reset(input); assign(output,'culture.ans'); rewrite(output); readln(n,k,m,s,t);fillchar(f,sizeof(f),true);fillchar(v,sizeof(v),false); for i:=1 to n do read(wi); for i:=1 to k do for j:=1 to k do read(ai,j); for i:=1 to m do begin readln(p,q,r); if bp,q=0 then begin bp,q:=r; bq,p:=r; end else if (r<bp,q)or(r<bq,p) then begin bp,q:=r; bq,p:=r; end; end; fs:=false; vws: