• Codeforces Round #261 (Div. 2)


    C.Pashmak and Buses

    构造

    看成k进制的d位数。。

     1 n,k,d = map(int, raw_input().split())
     2 if n > k**d: print -1
     3 else:
     4     ans = []
     5     for i in range(n):
     6         tmp = i
     7         cur = []
     8         for j in range(d):
     9             cur.append(tmp%k+1)
    10             tmp /= k
    11         ans.append(cur)
    12     for i in range(d):
    13         for j in ans:
    14             print j[i],
    15         print ''
    View Code

    D.http://codeforces.com/contest/459/problem/D

    树状数组求逆序数

    没怎么看懂思路。。噗。。

    参考:http://www.cnblogs.com/yuiffy/p/3916512.html

    E.http://codeforces.com/contest/459/problem/E

     1 /*Author :usedrose  */
     2 /*Created Time :2015/8/9 21:56:08*/
     3 /*File Name :2.cpp*/
     4 #pragma comment(linker, "/STACK:1024000000,1024000000") 
     5 #include <cstdio>
     6 #include <iostream>
     7 #include <algorithm>
     8 #include <sstream>
     9 #include <cstdlib>
    10 #include <cstring>
    11 #include <climits>
    12 #include <vector>
    13 #include <string>
    14 #include <ctime>
    15 #include <cmath>
    16 #include <deque>
    17 #include <queue>
    18 #include <stack>
    19 #include <set>
    20 #include <map>
    21 #define INF 0x3f3f3f3f
    22 #define eps 1e-8
    23 #define pi acos(-1.0)
    24 #define MAXN 1110
    25 #define MAXM 300110
    26 #define OK cout << "ok" << endl;
    27 #define o(a) cout << #a << " = " << a << endl
    28 #define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
    29 using namespace std;
    30 typedef long long LL;
    31 
    32 struct node {
    33     int u, v, w;
    34     bool operator<(const node& ss) const {
    35         return w < ss.w;
    36     }
    37 }e[MAXM];
    38 int n, m;
    39 int dp[MAXM], g[MAXM];
    40 
    41 int main()
    42 {
    43     cin.tie(0);
    44     ios::sync_with_stdio(false);
    45     cin >> n >> m;
    46     for (int i = 0;i < m; ++ i) 
    47         cin >> e[i].u >> e[i].v >> e[i].w;
    48     sort(e, e + m);
    49     int t = 0, ans = 0;
    50     for (int i = 0;i < m; ++ i) {
    51         dp[i] = g[e[i].u] + 1;
    52         if (e[i].w != e[i+1].w) {
    53             for (int j = t; j <= i; ++ j)
    54                 g[e[j].v] = max(dp[j], g[e[j].v]);
    55             t = i+1;
    56         }
    57         ans = max(ans, dp[i]);
    58     }
    59     cout << ans << endl;
    60     return 0;
    61 }
    View Code

    参考

    http://blog.csdn.net/y990041769/article/details/38709519

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  • 原文地址:https://www.cnblogs.com/usedrosee/p/4714952.html
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