[noip2014]解方程 hash+秦九昭

坑啊= =

选了好几次质数，发现还是这一组靠谱

思路：每次mod之后求出所有解，再mod再求，看着复杂度差不多了就把没重复的都输出就行了

const mi:array[1..7] of int64=(12537,15437,17647,14677,10003,10009,10007);
var
n,m,shi,sum,x:int64;
i,j,k,y1:longint;
a:array[-1..102,-1..10000+9] of int64;
flag:array[-1..1000000+9] of boolean;
c:char;
b:array[-1..102,1..7] of int64;
f:array[-1..100000,1..7] of boolean;

begin
readln(n,m);
fillchar(flag,sizeof(flag),true);
fillchar(f,sizeof(f),true);
for i:=0 to n do
begin
read(c);
if c='-' then
begin j:=0; y1:=-1; end else begin y1:=1; j:=1; val(c,a[i,1],k); end;
a[i,1]:=a[i,1]*y1;
while not eoln do
begin
inc(j);
read(c);
val(c,a[i,j],k);
a[i,j]:=a[i,j]*y1;
end;
a[i,0]:=j;
readln;
end;
for k:=1 to 7 do
for i:=0 to n do
begin
shi:=1;
for j:=a[i,0] downto 1 do
begin
b[i,k]:=(b[i,k]+a[i,j]*shi mod mi[k]) mod mi[k];
shi:=shi*10 mod mi[k];
end;
end;
for k:=1 to 7 do
for i:=1 to mi[k] do
begin
sum:=0; x:=1;
for j:=0 to n do
begin
sum:=(sum+x*b[j,k] mod mi[k]) mod mi[k];
x:=x*i mod mi[k];
end;
if sum<>0 then f[i,k]:=false;
end;
sum:=0;
for i:=1 to m do
for k:=1 to 7 do
if not f[i mod mi[k],k] then begin flag[i]:=false; break; end;
for i:=1 to m do if flag[i] then inc(sum);
writeln(sum);
for i:=1 to m do if flag[i] then writeln(i);
end.

想学会这道题的同学们一定要自己打一遍，很重要23333

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