Yet another end of the world
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <cstdlib> 6 #include <algorithm> 7 #include <vector> 8 #include <stack> 9 #include <queue> 10 #include <cassert> 11 #include <set> 12 #include <sstream> 13 #include <map> 14 #include <bitset> 15 using namespace std ; 16 #define zero {0} 17 #define INF 2000000000 18 #define eps 1e-6 19 #define maxn 1005 20 int x[maxn]; 21 int y[maxn]; 22 int z[maxn]; 23 24 bool is(int i,int j) 25 { 26 int r,l; 27 if(y[i]>z[j]||y[j]>z[i]) 28 { 29 if(y[j]>z[i]) 30 swap(i,j); 31 r=z[i]-y[j]; 32 l=y[i]-z[j]; 33 int xx=__gcd(x[i],x[j]); 34 int k=1; 35 if((l/xx)*xx<=r&&(l/xx)*xx>=l) 36 return true; 37 if((l/xx+1)*xx>=l&&(l/xx+1)*xx<=r) 38 return true; 39 return false; 40 } 41 else 42 return true; 43 } 44 int main() 45 { 46 #ifdef DeBUG 47 freopen("C:\Users\Sky\Desktop\1.in","r",stdin); 48 #endif 49 int n; 50 while(scanf("%d",&n)+1) 51 { 52 int i,j,k; 53 int _flag=1; 54 for(i=0;i<n;i++) 55 scanf("%d %d %d",&x[i],&y[i],&z[i]); 56 for(i=0;i<n;i++) 57 { 58 for(j=i+1;j<n;j++) 59 { 60 if(is(i,j)) 61 { 62 _flag=0; 63 break; 64 } 65 } 66 } 67 if(_flag) 68 printf("Can Take off "); 69 else 70 printf("Cannot Take off "); 71 } 72 return 0; 73 }
此题印证了数学思路的重要性，数学题只要方向对了代码没几行的，判断gcd(x1,x2)的倍数在区间里可以O(1)的，结果SB的TLE三次