The Ball And Cups
At the end of a busy day, The Chef and his assistants play a game together. The game is not just for fun but also used to decide who will have to clean the kitchen. The Chef is a Game Master, so his concern is how to manage the game but not how to win the game like his assistants do.
The game requires players to find the only ball under one of the N cups after their positions are changed in a special way. At the beginning of the game, The Chef places N cups in a row and put a ball under the C-th cup from the left (the cups are numbered from 1 to N). All players can see the initial position of the ball. Then Chef performs Q flip operations. Each flip operation is defined by two integers L and R such that 1 ≤ L ≤ R ≤ N and consists in reversing the segment [L, R] of cups. Namely, Chef swaps L-th and R-th cups, (L+1)-th and (R−1)-th cups, and so on. After performing all the operations Chef asks his assistants to choose a cup that they think the ball is under it. Who can guess the position of the ball will win the game, and of course, the others will have to clean the kitchen.
The Chef doesn't want to check all the N cups at the end of the game. He notes down the value of C and the pairs (L, R) and asked you, the mastered programmer, to determine the cup that contains the ball.
Input
The first line of the input contains a single integer T, denoting the number of test cases. The description of Ttest cases follows. The first line of each test case contains three space-separated integers N, C and Q, denoting the total number of cups, the initial position of the ball and the number of flip operations Chef will perform. Each of the following Q lines contains two space-separated integers L and R, denoting the ends of the segment of the current flip operation.
Output
For each test case output on a separate line the final position of the ball.
Constraints
- 1 ≤ T ≤ 10
- 1 ≤ N ≤ 100000 (105)
- 1 ≤ C ≤ N
- 1 ≤ Q ≤ 10000 (104)
- 1 ≤ L ≤ R ≤ N
Example
Input: 1 5 2 3 1 4 3 5 1 5 Output: 1
也是個構造數學公式的樣例。
這裡是過萬個輸入。故此最優點理一下輸入，使得程式能夠0ms過。
注意：
1 陷阱 - C會不在[L, R]範圍內
2 fread處理輸入，記得推斷最後輸入結束的條件 - fread返回長度為零，否則。儘管能夠AC。可是程式是有bug的。
我都使用類當做函式使用了，能夠非常好降低變數名的衝突。
#pragma once
#include <stdio.h>
class TheBallAndCups
{
int st, len;
static const int BU_MAX = 5120;
char buffer[BU_MAX];
char getFromBuffer()
{
if (st >= len)
{
len = fread(buffer, 1, BU_MAX, stdin);
st = 0;
}
return buffer[st++];
}
int scanInt()
{
char c = getFromBuffer();
while (c < '0' || '9' < c)
{
c = getFromBuffer();
}
int num = 0;
while ('0' <= c && c <= '9' && 0 != len)//必需要加0 != len推斷輸入結束
{
num = (num<<3) + (num<<1) + (c - '0');
c = getFromBuffer();
}
return num;
}
public:
TheBallAndCups() : st(0), len(0)
{
int T = 0, N = 0, C = 0, L = 0, R = 0, Q = 0;
T = scanInt();
while (T--)
{
N = scanInt();
C = scanInt();
Q = scanInt();
while (Q--)
{
L = scanInt();
R = scanInt();
if (C < L || R < C) continue;
int M = L + ((R-L)>>1);
if (C <= M)
{
int diff = C - L;
C = R - diff;
}
else
{
int diff = R - C;
C = L + diff;
}
}
printf("%d\n", C);
}
}
};
int theBallAndCups()
{
TheBallAndCups();
return 0;
}