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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#417577#8723. 乘二skip2004ML 0ms0kbPascal1.2kb2024-05-22 19:50:112024-05-22 19:50:11

Judging History

你现在查看的是最新测评结果

  • [2024-05-22 19:50:11]
  • 评测
  • 测评结果:ML
  • 用时:0ms
  • 内存:0kb
  • [2024-05-22 19:50:11]
  • 提交

answer

const P = 1000000007;
var n, k, i, rem, tmp : longint;
a : array [1..200000] of longint;
ans, pw : int64;
procedure qsort(l,r:longint);
var i,j,x,y:longint;
begin
	i:=l; j:=r; x:=a[(l+r) div 2];
	repeat
		while (a[i]<x) and (i<j) do inc(i);
		while (a[j]>x) and (i<j) do dec(j);
		if (i<j) then
		begin
			y:=a[i];
			a[i]:=a[j];
			a[j]:=y;
			inc(i);
			dec(j);
		end;
	until i>=j;
	if (i<r) then qsort(i,r);
	if (j>l) then qsort(j,l);
end;
begin
	read(n, k);
	for i := 1 to n do
		read(a[i]);
	qsort(1, n);
	while (a[1] < P) and (k > 0) do
	begin
		tmp := a[1] * 2;
		for i := 2 to n do
			if a[i-1] > a[i] then
			begin
				writeln(-1);
				halt;
			end;


			for i := 1 to n do
			begin
				if (k = 0) or (a[i] > tmp) then
					break;
				a[i] := a[i] * 2;
				dec(k);
			end;
			qsort(1, n);
		end;
		pw := 1;
		rem := k mod n;
		k := k div n;
		while k > 30 do
		begin
			pw := (pw << 30) mod P;
			k := k - 30;
		end;
		for i := 1 to k do
			pw := pw * 2 mod P;

		for i := 1 to rem do
			ans := (ans + a[i] * pw * 2) mod P;
		for i := rem + 1 to n do
			ans := (ans + a[i] * pw) mod P;
		writeln(ans);
end.

詳細信息

Test #1:

score: 0
Memory Limit Exceeded

input:

3 3
7 2 1

output:


result: