[09 noip simulation] Light

From my Baidu Space

Light. PAS/C/CPP)

[Description]

On an infinitely long road, there is a line of infinite street lights numbered 1, 2, 3, 4 ,.......

Each lamp has only two possible states: on or off. If you click the switch of a certain lamp, the status of the lamp will change. If it is turned on, it will turn off. If it is disabled, it will be turned on.

At the beginning, all the lights were turned off.

James can perform the following operations each time:

Specify two numbers, a and T (A is a real number and T is a positive integer ). Enter [a], [2 * A], [3 * A],…, The switch of the [T * A] lamp is one by one. [K] indicates the integer part of the real number K.

After James performed N operations, James suddenly found that only one lamp was on at this time, and James wanted to know the number of the lamp, but this light is too far away from Xiaoming, and James cannot see the number.

Fortunately, James remembers n previous operations. So James found you. Could you help him calculate the number of the light on?

[Input format]

The first row is a positive integer N, indicating N operations.

Next there are n rows, two numbers in each row, AI and Ti. Where AI is a real number, there must be 6 digits after the decimal point, Ti is a positive integer.

[Output format]

Only one positive integer, the number of the light on.

[Input example]

3

1.618034 13

2.618034 7

1.000000 21

[Output example]

20

[Data scale]

Note T = t1 + T2 + T3 + ...... + Tn.

For 30% of data, T <= 1000

For 80% of data, T <= 200000

For 100% of data, T <= 2000000

For 100% of data, n <=, 1 <= AI <, 1 <= Ti <= T

Data guarantee: After N operations, only one light is on and no error is required.

The following is a program provided by myself. It is just a simulation. If you want Daniel to see it, you can instruct the younger brother how to optimize it ~

Program light (input, output );
VaR
I, i1, j, k, n, T1, Z1: longint;
Z: real;
A: array [1 .. 10000000] of real;
B: array [1 .. 10000000] of Boolean;
C: array [1 .. 1000] of integer;
T: array [1 .. 10000000] of integer;

Function Change (Z: Real): longint;
VaR I1: longint;
Begin
I1: = 0;
While I1 <Z do Inc (I1 );
Change: = I1;
End; {change}

Begin
Assign (input, 'light. in ');
Reset (input );
Assign (output, 'light. out ');
Rewrite (output );

Readln (N );
For I: = 1 to n do
Readln (A [I], t [I]);
Close (input );

K: = 1;
For I: = 1 to n do writeln (A [I], '', t [I]);
Fillchar (B, sizeof (B), false );
For I: = 1 to n do begin
For J: = 1 to T [I] Do begin
Z: = int (A [I] * j );
Z1: = change (z );
Write ('deng ----> ', Z1 ,'');
If Z1> K then K: = Z1;
B [Z1]: = Not (B [Z1]);
For I1: = 1 to k do write (B [I1]: 8 );
Writeln;
End;
End; {doit}
T1: = 0;
For I: = 1 to k do
If B [I] Then begin
T1: = t1 + 1;
C [T1]: = I;
End;
For I: = 1 to T1 do writeln (C [I]);
Close (output );
End.

Some test outputs are not removed

There were only seven points on the left, and later I found that it seemed to be simplified, and the following program passed 10 points.
Program light (input, output );
VaR
I, i1, j, k, n, T1, Z1: longint;
Z: real;
A: array [1 .. 200000000] of real;
B: array [1 .. 200000000] of Boolean;
C: array [1 .. 1000] of integer;
T: array [1 .. 200000000] of integer;

Function Change (Z: Real): longint;
VaR I1: longint;
Begin
I1: = 0;
While I1 <Z do Inc (I1 );
Change: = I1;
End; {change}

Begin
Assign (input, 'light. in ');
Reset (input );
Assign (output, 'light. out ');
Rewrite (output );

Readln (N );
For I: = 1 to n do
Readln (A [I], t [I]);
Close (input );

K: = 1;
Fillchar (B, sizeof (B), false );
For I: = 1 to n do begin
For J: = 1 to T [I] Do begin
Z: = int (A [I] * j );
Z1: = change (z );
If Z1> K then K: = Z1;
B [Z1]: = Not (B [Z1]);
End;
End; {doit}
T1: = 0;
For I: = 1 to k do
If B [I] Then begin
T1: = t1 + 1;
C [T1]: = I;
End;
For I: = 1 to T1 do writeln (C [I]);
Close (output );
End.