1. （子矩阵）输入一个 n1*m1 的矩阵 a，和 n2*m2 的矩阵 b，问 a 中是否存在子矩阵和 b 相等。 若存在， 输出所有子矩阵左上角的坐标； 若不存在输出 “There is no answer” 。
Const SIZE = 50;
Var
n1, m1, n2, m2, i, j, k1, k2 : Integer;
a, b : Array[1..SIZE, 1..SIZE] Of Integer;
good, haveAns : Boolean;
Begin Readln(n1, m1);
For i := 1 To n1 Do
For j := 1 To m1 Do
Read(a[i][j]);
Readln(n2, m2);
For i := 1 To n2 Do
For j := 1 To m2 Do
read(b[i][j]) ;
haveAns := FALSE;
For i := 1 To n1 - n2 + 1 Do
For j := 1 To m1-m2+1 Do
Begin
good:=true ;
For k1 := 1 To n2 Do
For k2 := 1 To m2 Do
If a[i + k1 - 1][j + k2 - 1] <> b[k1][k2] Then
good := FALSE;
If good Then
Begin
Writeln(i, ' ', j);
haveAns:=true
End;
End;
If NOT haveAns Then
Writeln('There is no answer');
End.
2. 大整数开方）输入一个正整数 n（1≤n＜10 ），试用二分法计算它的平方根的整数 部分。
Const SIZE = 200;
Type
hugeint = Record
len : Integer;
num : Array[1..SIZE] Of Integer;
End;
//len 表示大整数的位数;num[1]表示个位、num[2]表示十位,以此类推
Var
s : String;
i : Integer;
target, left, middle, right : hugeint;
Function times(a, b : hugeint) : hugeint;
// 计算大整数 a 和 b 的乘积
Var
i, j : Integer;
ans : hugeint;
Begin
FillChar(ans, SizeOf(ans), 0);
For i := 1 To a.len Do
For j := 1 To b.len Do
ans.num[i+j-1] := ans.num[i + j - 1] + a.num[i] * b.num[j];
For i := 1 To a.len + b.len Do
Begin
ans.num[i + 1] := ans.num[i + 1] + ans.num[i] DIV 10;
ans.num[i]:= ans.num[i] mod 10 ;
If ans.num[a.len + b.len] > 0
Then ans.len := a.len + b.len
Else ans.len := a.len + b.len - 1;
End;
times := ans;
End;
Function add(a, b : hugeint) : hugeint;
// 计算大整数 a 和 b 的和
Var
i : Integer;
ans : hugeint;
Begin
FillChar(ans.num, SizeOf(ans.num), 0);
If a.len > b.len
Then ans.len := a.len
Else ans.len := b.len;
For i := 1 To ans.len Do
Begin
ans.num[i] := ans.num[i]+a.num[i]+b.num[i] ;
ans.num[i + 1] := ans.num[i + 1] + ans.num[i] DIV 10;
ans.num[i] := ans.num[i] MOD 10;
End;
If ans.num[ans.len + 1] > 0
Then Inc(ans.len);
add := ans;
End;
Function average(a, b : hugeint) : hugeint;
// 计算大整数 a 和 b 的平均数的整数部分
Var
i : Integer;
ans : hugeint;
Begin
ans := add(a, b);
For i := ans.len DownTo 2 Do
Begin
ans.num[i - 1] := ans.num[i - 1] + ( ans.num[i] mod 2) )*10;
ans.num[i] := ans.num[i] DIV 2;
End;
ans.num[1] := ans.num[1] DIV 2;
If ans.num[ans.len] = 0
Then Dec(ans.len);
average := ans;
End;
Function plustwo(a : hugeint) : hugeint;
// 计算大整数 a 加 2 后的结果
Var
i : Integer;
ans : hugeint;
Begin ans := a;
ans.num[1] := ans.num[1] + 2;
i := 1;
While (i <= ans.len) AND (ans.num[i] >= 10) Do Begin ans.num[i + 1] := ans.num[i + 1] + ans.num[i] DIV 10;
ans.num[i] := ans.num[i] MOD 10;
Inc(i);
End;
If ans.num[ans.len + 1] > 0
Then inc(ans.len) ;
plustwo := ans;
End;
Function over(a, b : hugeint) : Boolean;
// 若大整数 a > b 则返回 1, 否则返回 0
Var
i : Integer;
Begin
If ( a.len<b.len ) Then
Begin over := FALSE;
Exit;
End;
If a.len > b.len Then
Begin
over := TRUE;
Exit;
End;
For i := a.len DownTo 1 Do
Begin
If a.num[i] < b.num[i] Then
Begin
over := FALSE;
Exit;
End;
If a.num[i] > b.num[i] Then
Begin
over := TRUE;
Exit;
End;
End;
over := FALSE;
End;
Begin Readln(s);
FillChar(target.num, SizeOf(target.num), 0);
target.len := Length(s);
For i := 1 To target.len Do
target.num[i] := Ord(s[target.len - i + 1]) -ord('0') ;
FillChar(left.num, SizeOf(left.num), 0);
left.len := 1;
left.num[1] := 1;
right := target;
Repeat
middle := average(left, right);
If over( times(middle, middle),target )
Then right := middle
Else left := middle;
Until over(plustwo(left), right);
For i := left.len DownTo 1 Do
Write(left.num[i]);
Writeln;
End.