Codeforces Round #427 (Div. 2)——ABCD

http://codeforces.com/contest/835

A.拼英语水平和手速的签到题

 #include <bits/stdc++.h>

 using namespace std;

 int s, v1, v2, t1, t2;

 int main() {
     cin >> s >> v1 >> v2 >> t1 >> t2;
     int a1 = t1 + v1 * s + t1;
     int a2 = t2 + v2 * s + t2;
     if(a1 < a2) puts("First");
     else if(a1 > a2) puts("Second");
     else puts("Friendship");
     return ;
 }

B.同签到

 #include <bits/stdc++.h>

 using namespace std;

 int k, n, a[];

 string s;

 long long sum;

 int main() {
     cin >> k >> s;
     for(int i = ;i < s.size();i ++)
         a[s[i] - ''] ++, sum += s[i] - '';
     if(sum >= k) {
         puts("");
         return ;
     }
     k -= sum;
     for(int i = ;i < ;i ++)
         for(int j = ;j <= a[i];j ++) {
             k -=  - i, n ++;
             if(k <= ) {
                 printf("%d\n", n);
                 return ;
             }
         }
     return ;
 }

C.看到xyc的数据范围就能明白是道水题...然后就WA了

1W个位置10W个点，可能一个位置好几颗星星...令人无语

 #include <bits/stdc++.h>

 using namespace std;

 vector<int> c[][][];

 int n, q, cc, b[][][];

 int main() {
     scanf("%d %d %d", &n, &q, &cc), cc ++;
     for(int u, v, w, i = ;i <= n;i ++) {
         scanf("%d %d %d", &u, &v, &w);
         c[][u][v].push_back(w);
     }
     for(int i = ;i < cc;i ++)
         for(int j = ;j <= ;j ++)
             for(int k = ;k <= ;k ++)
                 if(c[i - ][j][k].size() != ) {
                     for(int p = ;p < c[i - ][j][k].size();p ++)
                         c[i][j][k].push_back((c[i - ][j][k][p] + ) % cc);
                 }
     for(int i = ;i < cc;i ++)
         for(int j = ;j <= ;j ++)
             for(int k = ;k <= ;k ++) {
                 b[i][j][k] += b[i][j - ][k] + b[i][j][k - ] - b[i][j - ][k - ];
                 for(int p = ;p < c[i][j][k].size();p ++)
                     b[i][j][k] += c[i][j][k][p];
             }
     for(int t, A, B, C, D, i = ;i <= q;i ++) {
         scanf("%d %d %d %d %d", &t, &A, &B, &C, &D);
         t %= cc, printf("%d\n", b[t][C][D] + b[t][A - ][B - ] - b[t][A - ][D] - b[t][C][B - ]);
     }
     return ;
 }

如果没有亮度变化，坐标到1e9，会有新的星星在某个位置出现，询问不变

允许离线的话可以CDQ分治解决，O(nlog^2n)

强制在线的话可以树状数组套主席树，时间复杂度相同，常数略大，空间O(nlogn)

当然这个模型感觉都快被写烂了...

D.注意到1阶回文是普通回文

而k阶回文决定了它一定也是回文，而且它也是1...k-1阶回文

想清楚了它的性质，就是个区间DP了

 import java.util.Scanner;

 public class D {
     public static void main(String []args) {
         Scanner cin = new Scanner(System.in);
         int[][] dp = new int[5005][5005];
         int[] ans = new int[5005];
         String s = cin.nextLine();
         int n = s.length();
         for(int d = 1;d <= n;d ++)
             for(int i = 0;i + d - 1< n;i ++) {
                 int j = i + d - 1;
                 if(d < 3) dp[i][j] = (s.charAt(i) == s.charAt(j) ? d : 0);
                 else if(s.charAt(i) != s.charAt(j) || dp[i + 1][j - 1] == 0) dp[i][j] = 0;
                 else if(s.charAt(i) == s.charAt((i + j - 1) / 2)) dp[i][j] = dp[i][(i + j - 1) / 2] + 1;
                 else dp[i][j] = 1;
                 ans[dp[i][j]] ++;
             }
         for(int i = n - 1;i > 0;i --)
             ans[i] += ans[i + 1];
         for(int i = 1;i <= n;i ++)
             System.out.printf("%d ", ans[i]);
     }
 }