noipday3提高组模拟题————解题报告
2015-10-20 20:57
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1.方程式
思路：直接枚举根，然后通过逆向多项式乘法提出这个根因式。
参考程序：
var
t,n,i,j:longint;
a,b:array[0..11000]of longint;
function calc(k:longint):longint;
var
i,tt:longint;
begin
calc:=0;tt:=1;
for i:=1 to n do
begin
calc:=calc+a[i]*tt;
// if k=1 then writeln(calc);
tt:=tt*k;
end;
end;
procedure change(k:longint);
var
i:longint;
begin
b[1]:=a[1] div (-k);
for i:=2 to n-1 do
b[i]:=(b[i-1]-a[i])div k;
n:=n-1;
a:=b;
end;
begin
assign(input,'equation.in');reset(input);
assign(output,'equation.out');rewrite(output);
read(n);
n:=n+1;
for i:=1 to n do read(a[i]);
//for i:=1 to n shr 1 do begin t:=a[i];a[i]:=a[n-i+1];a[n-i+1]:=t;end;
for i:=1 to 20 do
while calc(i)=0 do
begin
write(i,' ');
change(i);
//for j:=1 to n do write(a[j],' ');writeln;
end;
close(input);close(output);
end.
2.高维宇宙
思路：标程是状压dp，我想的是二分图匹配。
此题中质数只能是由奇数加偶数，因此可将一般图弄为二分图，网络流即可。
参考程序（标准的dinic算法,只有90分，据目测还有一个数据可能有问题）：
type
edge=record
tt,v,next:longint;
end;
var
e:array[0..110000]of edge;
b,level,head,q,a:array[0..110000]of longint;
cnt,cnt1,cnt2,n:longint;
procedure openfile;
begin
assign(input,'prime.in');reset(input);
assign(output,'prime.out');rewrite(output);
end;
procedure closfile;
begin
close(input);close(output);
end;
procedure printf(k:longint);
begin
write(k);
end;
function min(a,b:longint):longint;
begin
if a<b then exit(a);
exit(b);
end;
procedure init;
var
i,x:longint;
begin
readln(n);
for i:=1 to n do
begin
read(x);
if x and 1=1 then begin cnt1:=cnt1+1;a[cnt1]:=x;end
else begin cnt2:=cnt2+1;b[cnt2]:=x;end;
end;
end;
procedure insert(u,v,w:longint);
begin
cnt:=cnt+1;
e[cnt].tt:=v;
e[cnt].v:=w;
e[cnt].next:=head[u];
head[u]:=cnt;
end;
procedure build;
var
ok:array[0..2200]of boolean;
i,j:longint;
begin
ok[1]:=true;
for i:=2 to trunc(sqrt(2100)) do//第二版：：大家注意到了么？去掉之后就是。。。。Accepted......
if not ok[i] then
for j:=2 to 2100 div i do
ok[i*j]:=true;
for i:=1 to cnt1 do
for j:=1 to cnt2 do
if not ok[a[i]+b[j]] then
begin
//writeln(i,' ',j+cnt1);
insert(i,j+cnt1,1);
insert(j+cnt1,i,0);
end;
for i:=1 to cnt1 do begin {writeln(0,' ',i);}insert(0,i,1);insert(i,0,0);end;
for i:=1 to cnt2 do begin {writeln(i,' ',n+1);}insert(i,n+1,1);insert(n+1,i,0);end;
end;
function bfs:boolean;
var
h,t,i,now:longint;
begin
fillchar(level,sizeof(level),$ff);
h:=0;t:=1;q[1]:=0;level[0]:=0;
while h<t do
begin
h:=h+1;
now:=q[h];
i:=head[now];
while i>0 do
begin
if (e[i].v>0)and(level[e[i].tt]=-1) then
begin
t:=t+1;
q[t]:=e[i].tt;
level[e[i].tt]:=level[now]+1;
end;
i:=e[i].next;
end;
end;
if level[n+1]=-1 then exit(false);
exit(true);
end;
function dfs(x,f:longint):longint;
var
i,used,w:longint;
begin
if x=n+1 then exit(f);
used:=0;i:=head[x];
while i>0 do
begin
if (e[i].v>0)and(level[e[i].tt]=level[x]+1) then
begin
w:=f-used;
w:=dfs(e[i].tt,min(e[i].v,w));
e[i].v:=e[i].v-w;
e[i xor 1].v:=e[i x
4000
or 1].v+w;
used:=used+w;
if used=f then exit(f);
end;
i:=e[i].next;
end;
if used=0 then level[x]:=-1;
exit(used);
end;
procedure dinic;
const
INF=maxlongint shr 2;
var
maxflow:longint;
begin
maxflow:=0;
while bfs do
maxflow:=maxflow+dfs(0,INF);
printf(maxflow);
end;
procedure solve;
begin
dinic;
end;
procedure main;
begin
openfile;
init;
build;
solve;
closfile;
end;
begin
main;
end.
第二版：：发现问题之所在————筛质数打错了，为什么要开根号呢？
此题可用状压dp写：
var
e:array[0..101000]of record
x,n:longint;
end;
l,t,nx,x,k,ans,b0,tt,a0,i,j,n:longint;
key:array[0..50,0..2001000]of boolean;
a,b:array[0..101000]of longint;
ok:array[0..22000]of boolean;
function min(a,b:longint):longint;
begin if a<b then exit(a);exit(b);end;
function max(a,b:longint):longint;
begin if a>b then exit(a);exit(b);end;
procedure build;
var
i,j:longint;
begin
ok[1]:=true;
fillchar(ok,sizeof(ok),0);
for i:=2 to 21000 do
if not ok[i] then
for j:=2 to 21000 div i do
ok[i*j]:=true;
end;
procedure add(x,y:longint);
begin
l:=l+1;//writeln(x,' ',y);
e[l].x:=y;e[l].n:=e[x].n;e[x].n:=l;
end;
function count(k:longint):longint;
begin
count:=0;
while k>0 do
begin
count:=count+k and 1;
k:=k shr 1;
end;
end;
begin
assign(input,'prime.in');reset(input);
assign(output,'prime.out');rewrite(output);
readln(n);
for i:=1 to n do
begin
read(tt);
if tt and 1=1 then begin a0:=a0+1;a[a0]:=tt;end
else begin b0:=b0+1;b[b0]:=tt;end;
end;
nx:=min(a0,b0);
if b0<a0 then
begin
for i:=0 to nx do
begin
t:=a[i];a[i]:=b[i];b[i]:=t;
end;
t:=a0;a0:=b0;b0:=t;
for i:=nx+1 to b0 do
b[i]:=a[i];
end;
l:=n;
build;
for i:=1 to b0 do
for j:=1 to a0 do
if not ok[b[i]+a[j]] then add(i,j);
key[0,0]:=true;
for i:=0 to b0-1 do
for j:=0 to 1<<nx-1 do
if key[i,j] then
begin
key[i+1,j]:=true;
x:=e[i+1].n;
while x>0 do
begin
k:=e[x].x-1;
key[i+1,j or (1<<k)]:=true;
x:=e[x].n;
end;
end;
ans:=0;
for j:=0 to 1<<nx-1 do
if key[b0,j] then ans:=max(ans,count(j));
write(ans);
close(input);close(output);
end.
3.麻将
思路：枚举打出的牌，再枚举要听的牌，深搜判断胡没胡即可。
参考程序：
var
{debug:array[0..1000]of record
no,num,flag:longint;
end;}
cnt:longint;
c,b,inq,x,a:array[0..1000]of longint;
procedure openfile;
begin
assign(input,'mahjong.in');reset(input);
assign(output,'mahjong.out');rewrite(output);
end;
procedure closfile;
begin
close(input);close(output);
end;
procedure init;
var
num,ch1,ch2:char;
i:longint;
begin
for i:=1 to 14 do
begin
read(num,ch1,ch2);
a[i]:=(ord(num)-48)*10+ord(ch1='p')+ord(ch1='w')*2;
x[a[i]]:=x[a[i]]+1;
end;
//s=0 p=1 w=2
end;
procedure prepare;
var
i:longint;
begin
for i:=1 to 9 do
begin
cnt:=cnt+1;
b[cnt]:=i*10;
b[cnt+1]:=i*10+1;
b[cnt+2]:=i*10+2;
cnt:=cnt+2;
end;
end;
{procedure print;
var
i:longint;
begin
for i:=1 to 4 do with debug[i] do writeln(no,' ',num,' ',flag);
for i:=1 to 14 do write(inq[c[i]],' ');writeln;
end;}
function dfs(last,step:longint):boolean;
var
i,u:longint;
bool:boolean;
begin
if last=2 then
begin
for i:=1 to 14 do
if inq[c[i]]=2 then exit(true);
exit(false);
end;
bool:=false;
for i:=1 to 14 do
begin
if inq[c[i]]>2 then
begin
inq[c[i]]:=inq[c[i]]-3;
//debug[step].no:=i;debug[step].num:=c[i];debug[step].flag:=1;
bool:=bool or dfs(last-3,step+1);
inq[c[i]]:=inq[c[i]]+3;
end;//duizi
if bool then exit(true);
if (inq[c[i]]>0)and(inq[c[i]+10]>0)and(inq[c[i]+20]>0) then
begin
inq[c[i]]:=inq[c[i]]-1;inq[c[i]+10]:=inq[c[i]+10]-1;inq[c[i]+20]:=inq[c[i]+20]-1;
//debug[step].no:=i;debug[step].num:=c[i];debug[step].flag:=2;
bool:=bool or dfs(last-3,step+1);
inq[c[i]]:=inq[c[i]]+1;inq[c[i]+10]:=inq[c[i]+10]+1;inq[c[i]+20]:=inq[c[i]+20]+1;
end;//shunzi
if bool then exit(true);
end;
exit(false);
end;
function check(tno,k:longint):boolean;
var tlt:longint;
begin
inq:=x;c:=a;
inq[k]:=inq[k]+1;inq[a[tno]]:=inq[a[tno]]-1;
c[tno]:=k;
{if k=b[1] then
begin
for tlt:=1 to 14 do write(c[tlt]:4);writeln;
for tlt:=1 to 14 do write(inq[c[tlt]]:4);writeln;writeln;
end;}
exit(dfs(14,1));
end;
procedure solve;
var
max,tt,no,i,j,tlt:longint;
begin
max:=0;no:=0;
for i:=1 to 14 do
begin
//s=0 p=1 w=2
tt:=0;
for j:=1 to cnt do
if inq[b[j]]<4 then
if check(i,b[j]) then
begin
tt:=tt+5-inq[b[j]];
//if i=9 then writeln(b[j],' ',tt);
end;
if tt>max then begin no:=i;max:=tt;end;
end;
write(no,' ',max);
end;
procedure main;
begin
openfile;
init;
prepare;
solve;
closfile;
end;
begin
main;
end.
思路：直接枚举根，然后通过逆向多项式乘法提出这个根因式。
参考程序：
var
t,n,i,j:longint;
a,b:array[0..11000]of longint;
function calc(k:longint):longint;
var
i,tt:longint;
begin
calc:=0;tt:=1;
for i:=1 to n do
begin
calc:=calc+a[i]*tt;
// if k=1 then writeln(calc);
tt:=tt*k;
end;
end;
procedure change(k:longint);
var
i:longint;
begin
b[1]:=a[1] div (-k);
for i:=2 to n-1 do
b[i]:=(b[i-1]-a[i])div k;
n:=n-1;
a:=b;
end;
begin
assign(input,'equation.in');reset(input);
assign(output,'equation.out');rewrite(output);
read(n);
n:=n+1;
for i:=1 to n do read(a[i]);
//for i:=1 to n shr 1 do begin t:=a[i];a[i]:=a[n-i+1];a[n-i+1]:=t;end;
for i:=1 to 20 do
while calc(i)=0 do
begin
write(i,' ');
change(i);
//for j:=1 to n do write(a[j],' ');writeln;
end;
close(input);close(output);
end.
2.高维宇宙
思路：标程是状压dp，我想的是二分图匹配。
此题中质数只能是由奇数加偶数，因此可将一般图弄为二分图，网络流即可。
参考程序（标准的dinic算法,只有90分，据目测还有一个数据可能有问题）：
type
edge=record
tt,v,next:longint;
end;
var
e:array[0..110000]of edge;
b,level,head,q,a:array[0..110000]of longint;
cnt,cnt1,cnt2,n:longint;
procedure openfile;
begin
assign(input,'prime.in');reset(input);
assign(output,'prime.out');rewrite(output);
end;
procedure closfile;
begin
close(input);close(output);
end;
procedure printf(k:longint);
begin
write(k);
end;
function min(a,b:longint):longint;
begin
if a<b then exit(a);
exit(b);
end;
procedure init;
var
i,x:longint;
begin
readln(n);
for i:=1 to n do
begin
read(x);
if x and 1=1 then begin cnt1:=cnt1+1;a[cnt1]:=x;end
else begin cnt2:=cnt2+1;b[cnt2]:=x;end;
end;
end;
procedure insert(u,v,w:longint);
begin
cnt:=cnt+1;
e[cnt].tt:=v;
e[cnt].v:=w;
e[cnt].next:=head[u];
head[u]:=cnt;
end;
procedure build;
var
ok:array[0..2200]of boolean;
i,j:longint;
begin
ok[1]:=true;
for i:=2 to trunc(sqrt(2100)) do//第二版：：大家注意到了么？去掉之后就是。。。。Accepted......
if not ok[i] then
for j:=2 to 2100 div i do
ok[i*j]:=true;
for i:=1 to cnt1 do
for j:=1 to cnt2 do
if not ok[a[i]+b[j]] then
begin
//writeln(i,' ',j+cnt1);
insert(i,j+cnt1,1);
insert(j+cnt1,i,0);
end;
for i:=1 to cnt1 do begin {writeln(0,' ',i);}insert(0,i,1);insert(i,0,0);end;
for i:=1 to cnt2 do begin {writeln(i,' ',n+1);}insert(i,n+1,1);insert(n+1,i,0);end;
end;
function bfs:boolean;
var
h,t,i,now:longint;
begin
fillchar(level,sizeof(level),$ff);
h:=0;t:=1;q[1]:=0;level[0]:=0;
while h<t do
begin
h:=h+1;
now:=q[h];
i:=head[now];
while i>0 do
begin
if (e[i].v>0)and(level[e[i].tt]=-1) then
begin
t:=t+1;
q[t]:=e[i].tt;
level[e[i].tt]:=level[now]+1;
end;
i:=e[i].next;
end;
end;
if level[n+1]=-1 then exit(false);
exit(true);
end;
function dfs(x,f:longint):longint;
var
i,used,w:longint;
begin
if x=n+1 then exit(f);
used:=0;i:=head[x];
while i>0 do
begin
if (e[i].v>0)and(level[e[i].tt]=level[x]+1) then
begin
w:=f-used;
w:=dfs(e[i].tt,min(e[i].v,w));
e[i].v:=e[i].v-w;
e[i xor 1].v:=e[i x
4000
or 1].v+w;
used:=used+w;
if used=f then exit(f);
end;
i:=e[i].next;
end;
if used=0 then level[x]:=-1;
exit(used);
end;
procedure dinic;
const
INF=maxlongint shr 2;
var
maxflow:longint;
begin
maxflow:=0;
while bfs do
maxflow:=maxflow+dfs(0,INF);
printf(maxflow);
end;
procedure solve;
begin
dinic;
end;
procedure main;
begin
openfile;
init;
build;
solve;
closfile;
end;
begin
main;
end.
第二版：：发现问题之所在————筛质数打错了，为什么要开根号呢？
此题可用状压dp写：
var
e:array[0..101000]of record
x,n:longint;
end;
l,t,nx,x,k,ans,b0,tt,a0,i,j,n:longint;
key:array[0..50,0..2001000]of boolean;
a,b:array[0..101000]of longint;
ok:array[0..22000]of boolean;
function min(a,b:longint):longint;
begin if a<b then exit(a);exit(b);end;
function max(a,b:longint):longint;
begin if a>b then exit(a);exit(b);end;
procedure build;
var
i,j:longint;
begin
ok[1]:=true;
fillchar(ok,sizeof(ok),0);
for i:=2 to 21000 do
if not ok[i] then
for j:=2 to 21000 div i do
ok[i*j]:=true;
end;
procedure add(x,y:longint);
begin
l:=l+1;//writeln(x,' ',y);
e[l].x:=y;e[l].n:=e[x].n;e[x].n:=l;
end;
function count(k:longint):longint;
begin
count:=0;
while k>0 do
begin
count:=count+k and 1;
k:=k shr 1;
end;
end;
begin
assign(input,'prime.in');reset(input);
assign(output,'prime.out');rewrite(output);
readln(n);
for i:=1 to n do
begin
read(tt);
if tt and 1=1 then begin a0:=a0+1;a[a0]:=tt;end
else begin b0:=b0+1;b[b0]:=tt;end;
end;
nx:=min(a0,b0);
if b0<a0 then
begin
for i:=0 to nx do
begin
t:=a[i];a[i]:=b[i];b[i]:=t;
end;
t:=a0;a0:=b0;b0:=t;
for i:=nx+1 to b0 do
b[i]:=a[i];
end;
l:=n;
build;
for i:=1 to b0 do
for j:=1 to a0 do
if not ok[b[i]+a[j]] then add(i,j);
key[0,0]:=true;
for i:=0 to b0-1 do
for j:=0 to 1<<nx-1 do
if key[i,j] then
begin
key[i+1,j]:=true;
x:=e[i+1].n;
while x>0 do
begin
k:=e[x].x-1;
key[i+1,j or (1<<k)]:=true;
x:=e[x].n;
end;
end;
ans:=0;
for j:=0 to 1<<nx-1 do
if key[b0,j] then ans:=max(ans,count(j));
write(ans);
close(input);close(output);
end.
3.麻将
思路：枚举打出的牌，再枚举要听的牌，深搜判断胡没胡即可。
参考程序：
var
{debug:array[0..1000]of record
no,num,flag:longint;
end;}
cnt:longint;
c,b,inq,x,a:array[0..1000]of longint;
procedure openfile;
begin
assign(input,'mahjong.in');reset(input);
assign(output,'mahjong.out');rewrite(output);
end;
procedure closfile;
begin
close(input);close(output);
end;
procedure init;
var
num,ch1,ch2:char;
i:longint;
begin
for i:=1 to 14 do
begin
read(num,ch1,ch2);
a[i]:=(ord(num)-48)*10+ord(ch1='p')+ord(ch1='w')*2;
x[a[i]]:=x[a[i]]+1;
end;
//s=0 p=1 w=2
end;
procedure prepare;
var
i:longint;
begin
for i:=1 to 9 do
begin
cnt:=cnt+1;
b[cnt]:=i*10;
b[cnt+1]:=i*10+1;
b[cnt+2]:=i*10+2;
cnt:=cnt+2;
end;
end;
{procedure print;
var
i:longint;
begin
for i:=1 to 4 do with debug[i] do writeln(no,' ',num,' ',flag);
for i:=1 to 14 do write(inq[c[i]],' ');writeln;
end;}
function dfs(last,step:longint):boolean;
var
i,u:longint;
bool:boolean;
begin
if last=2 then
begin
for i:=1 to 14 do
if inq[c[i]]=2 then exit(true);
exit(false);
end;
bool:=false;
for i:=1 to 14 do
begin
if inq[c[i]]>2 then
begin
inq[c[i]]:=inq[c[i]]-3;
//debug[step].no:=i;debug[step].num:=c[i];debug[step].flag:=1;
bool:=bool or dfs(last-3,step+1);
inq[c[i]]:=inq[c[i]]+3;
end;//duizi
if bool then exit(true);
if (inq[c[i]]>0)and(inq[c[i]+10]>0)and(inq[c[i]+20]>0) then
begin
inq[c[i]]:=inq[c[i]]-1;inq[c[i]+10]:=inq[c[i]+10]-1;inq[c[i]+20]:=inq[c[i]+20]-1;
//debug[step].no:=i;debug[step].num:=c[i];debug[step].flag:=2;
bool:=bool or dfs(last-3,step+1);
inq[c[i]]:=inq[c[i]]+1;inq[c[i]+10]:=inq[c[i]+10]+1;inq[c[i]+20]:=inq[c[i]+20]+1;
end;//shunzi
if bool then exit(true);
end;
exit(false);
end;
function check(tno,k:longint):boolean;
var tlt:longint;
begin
inq:=x;c:=a;
inq[k]:=inq[k]+1;inq[a[tno]]:=inq[a[tno]]-1;
c[tno]:=k;
{if k=b[1] then
begin
for tlt:=1 to 14 do write(c[tlt]:4);writeln;
for tlt:=1 to 14 do write(inq[c[tlt]]:4);writeln;writeln;
end;}
exit(dfs(14,1));
end;
procedure solve;
var
max,tt,no,i,j,tlt:longint;
begin
max:=0;no:=0;
for i:=1 to 14 do
begin
//s=0 p=1 w=2
tt:=0;
for j:=1 to cnt do
if inq[b[j]]<4 then
if check(i,b[j]) then
begin
tt:=tt+5-inq[b[j]];
//if i=9 then writeln(b[j],' ',tt);
end;
if tt>max then begin no:=i;max:=tt;end;
end;
write(no,' ',max);
end;
procedure main;
begin
openfile;
init;
prepare;
solve;
closfile;
end;
begin
main;
end.
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