[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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