首先我们知道,对于所有种情况,我们可以将每一位可以放的
数的值加起来,所有位置的乘起来,等于的就是最后的答案,具体
为什么正确,可以根据乘法分配律来想一想。
那么对于所有不做要求的,快速幂直接算就行了,然后快排下,就知道
每个位置不放那些值,减掉后乘进去就行了。
/************************************************************** Problem: 2751 User: BLADEVIL Language: Pascal Result: Accepted Time:344 ms Memory:1008 kb ****************************************************************/ //By BLADEVIL const d39 =1000000007; var n, m, k :longint; a, b :array[0..100010] of longint; ans :int64; procedure swap(var a,b:longint); var c :longint; begin c:=a; a:=b; b:=c; end; procedure qs(low,high:longint); var i, j, xx, yy :longint; begin i:=low; j:=high; xx:=a[(i+j) div 2]; yy:=b[(i+j) div 2]; while i<j do begin while (a[i]<xx) or (a[i]=xx) and (b[i]<yy) do inc(i); while (a[j]>xx) or (a[j]=xx) and (b[j]>yy) do dec(j); if i<=j then begin swap(a[i],a[j]); swap(b[i],b[j]); inc(i); dec(j); end; end; if i<high then qs(i,high); if j>low then qs(low,j); end; procedure init; var i :longint; begin read(n,m,k); for i:=1 to k do read(a[i],b[i]); qs(1,k); end; function mi(a,b:int64):int64; var sum :int64; begin sum:=a; mi:=1; while b<>0 do begin if b mod 2=1 then mi:=mi*sum mod d39; sum:=sum*sum mod d39; b:=b div 2; end; end; procedure main; var i :longint; sum, x, y, z :int64; begin sum:=m; x:=-1; for i:=1 to k do begin if a[i]<>x then begin dec(sum); x:=a[i]; end; end; x:=n; y:=n+1; if x mod 2=0 then x:=x div 2 else y:=y div 2; x:=x mod d39; y:=y mod d39; x:=x*y mod d39; ans:=mi(x,sum); for i:=1 to k do if (a[i]=a[i-1]) and (b[i]=b[i-1]) then b[i-1]:=0; y:=-1; z:=-1; for i:=1 to k do begin if a[i]<>y then begin if i<>1 then ans:=ans*z mod d39; z:=x; y:=a[i]; z:=((x-b[i]) mod d39+d39) mod d39; end else z:=((z-b[i])mod d39+d39) mod d39; end; if z<>-1 then ans:=ans*z mod d39; writeln(ans); end; begin init; main; end.