POJ 1061

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题意:

　　两只青蛙在同一条纬度上,它们各自朝西跳,问它们要跳多少步才能碰面(必须同时到达同一点).

分析:

　　假设它们跳了t步才相遇,青蛙a初始坐标为x,青蛙b初始坐标为y,则跳了t步相遇后a的坐标为 x+m*t-p1*l, b的坐标为 y+n*t-p2*l (p1,p2分别表示a,b跳

的圈数) x+m*t-p1*l = y+n*t-p2*l => x-y = (n-m)*t + (p1-p2)*l => x-y = (n-m)*t + p*l.

代码如下:

 1 //方法一
 2 
 3 
 4 #include <iostream>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <fstream>
 8 #include <ctime>
 9 #include <cmath>
10 #include <cstdlib>
11 #include <algorithm>
12 #include <set>
13 #include <map>
14 #include <list>
15 #include <stack>
16 #include <queue>
17 #include <iterator>
18 #include <vector>
19 
20 using namespace std;
21 
22 #define LL long long
23 #define INF 0x3f3f3f3f
24 #define MOD 1000000007
25 #define MAXN 10000010
26 #define MAXM 1000010
27 
28 
29 LL extend_gcd(LL a, LL b, LL &x, LL &y)
30 {
31     if(b == 0)
32     {
33         x = 1;
34         y = 0;
35         return a;
36     }
37     else
38     {
39         LL r = extend_gcd(b, a%b, y, x);
40         y -= x*(a/b);
41         return r;
42     }
43 }
44 
45 
46 int main()
47 {
48     LL x, y, m, n, l;
49     while(scanf("%lld%lld%lld%lld%lld", &x, &y, &m, &n, &l)==5)
50     {
51         LL t, p;
52         LL ans = extend_gcd(n-m, l, t, p);
53         if((x-y)%ans)
54             printf("Impossible\n");
55         else
56         {
57             t = t*(x-y)/ans;
58             t = (t%(l/ans)+(l/ans))%(l/ans);
59             printf("%lld\n", t);
60         }
61     }
62 
63     return 0;
64 }

 1 //方法二
 2 
 3 
 4 #include <iostream>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <fstream>
 8 #include <ctime>
 9 #include <cmath>
10 #include <cstdlib>
11 #include <algorithm>
12 #include <set>
13 #include <map>
14 #include <list>
15 #include <stack>
16 #include <queue>
17 #include <iterator>
18 #include <vector>
19 
20 using namespace std;
21 
22 #define LL long long
23 #define INF 0x3f3f3f3f
24 #define MOD 1000000007
25 #define MAXN 10000010
26 #define MAXM 1000010
27 
28 
29 LL extend_gcd(LL a, LL b, LL &x, LL &y)
30 {
31     if(b == 0)
32     {
33         x = 1;
34         y = 0;
35         return a;
36     }
37     else
38     {
39         LL r = extend_gcd(b, a%b, y, x);
40         y -= x*(a/b);
41         return r;
42     }
43 }
44 
45 
46 LL cal(LL a, LL b, LL c)
47 {
48     LL x, y;
49     LL ans = extend_gcd(a, b, x, y);
50     if(c%ans)
51         return -1;
52     x *= c/ans;
53     b /= ans;
54     if(b < 0)
55         b = b*(-1);
56 //    b =  abs(b);  //abs只能对int取绝对值
57     ans = x%b;
58     if(ans <= 0)
59         ans += b;
60     return ans;
61 }
62 
63 int main()
64 {
65     LL x, y, m, n, l;
66     while(scanf("%lld%lld%lld%lld%lld", &x, &y, &m, &n, &l)==5)
67     {
68         LL sum = cal(n-m, l, x-y);
69         if(sum == -1)
70             printf("Impossible\n");
71         else
72             printf("%lld\n", sum);
73     }
74 
75     return 0;
76 }

posted on 2016-08-03 19:43 tony-cao 阅读(158) 评论(0) 收藏举报

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