Yet another end of the world

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 920    Accepted Submission(s):
400

Problem Description

In the year 3013, it has been 1000 years since the
previous predicted rapture. However, the Maya will not play a joke any more and
the Rapture finally comes in. Fortunately people have already found out
habitable planets, and made enough airships to convey all the human beings in
the world. A large amount of airships are flying away the earth. People all bear
to watch as this planet on which they have lived for millions of years.
Nonetheless, scientists are worrying about anther problem…
As we know that
long distance space travels are realized through the wormholes, which are given
birth by the distortion of the energy fields in space. Airships will be driven
into the wormholes to reach the other side of the universe by the suction
devices placed in advance. Each wormhole has its configured attract parameters,
X, Y or Z. When the value of ID%X is in [Y,Z], this spaceship will be sucked
into the wormhole by the huge attraction. However, the spaceship would be tear
into piece if its ID meets the attract parameters of two wormholes or more at
the same time.
All the parameters are carefully adjusted initially, but some
conservative, who treat the Rapture as a grain of truth and who are reluctant to
abandon the treasure, combine with some evil scientists and disrupt the
parameters. As a consequence, before the spaceships fly into gravity range, we
should know whether the great tragedy would happen or not. Now the mission is on
you.

Input

Multiple test cases, ends with EOF.
In each case,
the first line contains an integer N(N<=1000), which means the number of the
wormholes.
Then comes N lines, each line contains three integers
X,Y,Z(0<=Y<=Z<X<2*10⁹).

Output

If there exists danger, output “Cannot Take off”, else
output “Can Take off”.

Sample Input

2
7 2 3
7 5 6
2
7 2 2
9 2 2

Sample Output

Can Take off
Cannot Take off

题目大意：

　　这道题是说，给你n个x，y，z。问是否存在某个id，使得 id%x>=y&&id%x<=z

　　y <= z < 2*10^9

解题思路：

　　设  id/x1 = a1; id/x2 = a2;

　　　id%x1=b1; id%x2=a2;

-> id = x1*a1+b1,  id = x2*a2+b2

->x1*a1+b1 = x2*a2+b2;

->x1*a1-x2*a2 = b2-b1;

现在就来看看该怎么求解这个线性方程组了。首先，线性方程组有整数解的条件是b2-b1是gcd(x1,x2)的整数倍数。

那么，我们就可以得到只需要判断一下两个区间[yi,zi]和[yj,zj]的差值区间[yj-zi,zj-yi]是否可能存在gcd(xi,xj)的倍数的结论

代码：

# include<cstdio>
# include<iostream>

using namespace std;

# define MAX 1234

int x[MAX],y[MAX],z[MAX];
int n;

int gcd ( int a,int b )
{
    if ( b==0 )
        return a;
    else
        return gcd(b,a%b);
}

int ok ( int t,int l,int r )
{
    if ( l%t==0||r%t==0 )
        return 1;
    if ( l<0&&r>=0 )
        return 1;
    if ( r/t-l/t > 0 )
        return 1;
    return 0;
}

int solve()
{
    for ( int i = 1;i <= n;i++ )
    {
        for ( int j = i+1;j <= n;j++ )
        {
            int tmp = gcd(x[i],x[j]);
            if ( ok(tmp,y[i]-z[j],z[i]-y[j]) )
                return 1;
        }
    }
    return 0;
}

int main(void)
{
    while ( scanf("%d",&n)!=EOF )
    {
        for ( int i = 1;i <= n;i++ )
            scanf("%d%d%d",&x[i],&y[i],&z[i]);
        if ( solve() )
            puts("Cannot Take off");
        else
            puts("Can Take off");
    }

    return 0;
}