Codeforces Round #367 (Div. 2)

 1 #include <cstdio>
 2 #include <cmath>
 3 using namespace std;
 4 int a,b,n;
 5 struct _
 6 {
 7     int x,y,v;
 8 }p[1005];
 9 double dis2(_ A)
10 {
11     return (a - A.x) * (a - A.x) + (b - A.y) * (b - A.y);
12 }
13 int main()
14 {
15     scanf("%d%d%d", &a, &b, &n);
16     for(int i = 0; i < n; i++)
17         scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].v);
18     double ans = 1.0 * 0x3f3f3f3f;
19     for(int i = 0; i < n; i++)
20     {
21         double temp = (double) sqrt(dis2(p[i])) / p[i].v;
22         if(ans - temp > 1e-8)
23             ans = temp;
24     }
25     printf("%lf\n",ans);
26     return 0;
27 }

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 int a[100000 + 5];
 5 int main()
 6 {
 7     int n,q,x;
 8     scanf("%d", &n);
 9     for(int i = 0; i < n; i++)
10         scanf("%d", &a[i]);
11     scanf("%d", &q);
12     sort(a,a+n);
13     for(int i = 0; i < q; i++)
14     {
15         scanf("%d", &x);
16         int pos = upper_bound(a,a+n,x) - a;
17         printf("%d\n",pos);
18     }
19     return 0;
20 }

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 using namespace std;
 6 typedef long long LL;
 7 const int maxn = 100000 + 5;
 8 const LL INF = 1e17;
 9 int c[maxn];
10 LL dp[maxn][2];
11 string s[maxn];
12 int main()
13 {
14     int n;
15     cin>>n;
16     for(int i = 0; i < n; i++)
17         cin>>c[i];
18     for(int i = 0; i < n; i++)
19         cin>>s[i];
20     //init
21     for(int i = 0; i < n; i++)
22     {
23         dp[i][0] = INF;
24         dp[i][1] = INF;
25     }
26     dp[0][0] = 0;
27     dp[0][1] = c[0];
28     string s10,s11,s20,s21;
29     for(int i = 1; i < n; i++)
30     {
31         s10 = s11 = s[i-1];
32         reverse(s11.begin(),s11.end());
33 
34         s20 = s21 = s[i];
35         reverse(s21.begin(),s21.end());
36 
37         if(s10 <= s20)
38             dp[i][0] = min(dp[i][0], dp[i-1][0]);
39         if(s11 <= s20)
40             dp[i][0] = min(dp[i][0], dp[i-1][1]);
41         if(s10 <= s21)
42             dp[i][1] = min(dp[i][1], dp[i-1][0] + (LL)c[i]);
43         if(s11 <= s21)
44             dp[i][1] = min(dp[i][1], dp[i-1][1] + (LL)c[i]);
45     }
46     LL ans = min(dp[n-1][0], dp[n-1][1]);
47     cout<<(ans == INF ? -1 : ans)<<endl;
48 
49     return 0;
50 }