hdu 4596 Yet another end of the world （一阶不定方程可解性 ）

Yet another end of the world

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 265    Accepted Submission(s): 148

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*10 ⁹).

 

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
   
   
   
   

 

Source

2013 ACM-ICPC南京赛区全国邀请赛——题目重现

题意：

给出n组x，y，z。判断是否存在一个id使得id%x1∈（y1,z1），id%x2∈（y2,z2）。

思路：

设 id/x1=a , id/x2=b ,则

id-a*x1=u;   (1)

id-b*x2=v;   (2)

(1)-(2)得  一阶不定方程:  bx2-ax1=u-v .

方程可解性为是否存在 (u-v) 是gcd(x1,x2)的整数倍。

那么问题就解决了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 100005
#define mod 100000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans;
int x[maxn],y[maxn],z[maxn];

int gcd(int u,int v)
{
    if(v==0) return u;
    return gcd(v,u%v);
}
bool isok(int t,int le,int ri)
{
    int i,j;
    if(le%t==0||ri%t==0) return true ;
    if(le<0&&ri>=0) return true ;
    if(ri/t-le/t>0) return true ;
    return false ;
}
bool solve()
{
    int i,j,t,ma,mi;
    for(i=1;i<=n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            t=gcd(x[i],x[j]);
            if(isok(t,y[i]-z[j],z[i]-y[j])) return true ;
        }
    }
    return false ;
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&x[i],&y[i],&z[i]);
        }
        if(solve()) printf("Cannot Take off\n");
        else printf("Can Take off\n");
    }
    return 0;
}