随笔分类 -  算法竞赛

上一页 1 ··· 7 8 9 10 11 12 13 下一页
CodeForces 459D Pashmak and Parmida's problem
摘要：Pashmak and Parmida's problem Time Limit:3000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 459D Pashmak and 阅读全文
posted @ 2016-05-09 01:18 超级学渣渣 阅读(391) 评论(0) 推荐(0)
cf 459c Pashmak and Buses
摘要：E - Pashmak and Buses Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 459C E - Pashmak and Bus 阅读全文
posted @ 2016-05-08 19:46 超级学渣渣 阅读(227) 评论(0) 推荐(0)
hdu 1006 Tick and Tick 有技巧的暴力
摘要：Tick and Tick Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16707 Accepted Submission(s): 4083 阅读全文
posted @ 2016-05-08 18:44 超级学渣渣 阅读(717) 评论(0) 推荐(0)
hdu 1005 Number Sequence
摘要：Number Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 147906 Accepted Submission(s): 35 阅读全文
posted @ 2016-05-05 15:22 超级学渣渣 阅读(186) 评论(0) 推荐(0)
hdu 1004 Let the Balloon Rise
摘要：Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 102616 Accepted Submission(s 阅读全文
posted @ 2016-05-05 10:35 超级学渣渣 阅读(181) 评论(0) 推荐(0)
hdu1003 最大子串和
摘要：Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 207874 Accepted Submission(s): 48646 Prob 阅读全文
posted @ 2016-05-04 23:00 超级学渣渣 阅读(247) 评论(0) 推荐(0)
cf339d Xenia and Bit Operations
摘要：Xenia and Bit Operations Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 339D Xenia and Bit Op 阅读全文
posted @ 2016-05-04 21:05 超级学渣渣 阅读(390) 评论(0) 推荐(0)
A + B Problem II
摘要：A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 305856 Accepted Submission(s): 5 阅读全文
posted @ 2016-04-28 23:52 超级学渣渣 阅读(211) 评论(0) 推荐(0)
CodeForces 558D
摘要：Guess Your Way Out! II Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 558D Guess Your Way Out 阅读全文
posted @ 2016-04-26 16:11 超级学渣渣 阅读(319) 评论(1) 推荐(0)
Vanya and Brackets
摘要：Vanya and Brackets Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 552E Vanya and Brackets Sub 阅读全文
posted @ 2016-04-24 20:45 超级学渣渣 阅读(154) 评论(0) 推荐(0)
spfa
摘要：#include<iostream> #include<stdio.h> #include<queue> using namespace std; const int maxx = 105; const int INF = 99999999; int map[maxx][maxx],dist[max 阅读全文
posted @ 2016-04-24 11:37 超级学渣渣 阅读(126) 评论(0) 推荐(0)
hdu 1217 Arbitrage
摘要：Arbitrage Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6414 Accepted Submission(s): 2966 Probl 阅读全文
posted @ 2016-04-23 20:37 超级学渣渣 阅读(278) 评论(0) 推荐(0)
CodeForces 1A Theatre Square
摘要：A - Theatre Square Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 1A A - Theatre Square Submit 阅读全文
posted @ 2016-04-21 21:21 超级学渣渣 阅读(239) 评论(0) 推荐(0)
Shortest path of the king
摘要：Shortest path of the king time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output The king is left 阅读全文
posted @ 2016-04-21 21:16 超级学渣渣 阅读(272) 评论(0) 推荐(0)
map的用法
摘要：map感觉是非常方便的。 以后有其他的再更新。 阅读全文
posted @ 2016-04-21 19:56 超级学渣渣 阅读(170) 评论(0) 推荐(0)
CodeForces 2A Winner
摘要：Winner Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 2A Winner Submit Status Practice CodeFor 阅读全文
posted @ 2016-04-21 19:50 超级学渣渣 阅读(316) 评论(0) 推荐(0)
floyd算法&迪杰斯特拉算法
摘要：for(int k=1; k<=n; k++) for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) { gra[i][j]=min(gra[i][j],gra[i][k]+gra[k][j]); } void Dijkstra(int n, int v, 阅读全文
posted @ 2016-04-21 18:57 超级学渣渣 阅读(267) 评论(0) 推荐(0)
hdu 2112 HDU Today
摘要：HDU Today Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22939 Accepted Submission(s): 5485 Pro 阅读全文
posted @ 2016-04-21 18:54 超级学渣渣 阅读(203) 评论(0) 推荐(0)
CodeForces 306B
摘要：//#include<bits/stdc++.h> #include<algorithm> #include<stdio.h> #include<iostream> using namespace std; const int maxx=1000000*2+10; struct Node { int 阅读全文
posted @ 2016-04-20 09:02 超级学渣渣 阅读(298) 评论(0) 推荐(0)
同模余定理
摘要：声明：借鉴高手！ 一、 同余 对于整数除以某个正整数的问题，如果只关心余数的情况，就产生同余的概念。 定义1 用给定的正整数m分别除整数a、b，如果所得的余数相等，则称a、b对模m同余，记作a≡b(mod m)，如 56≡0 （mod 8）。 定理1 整数a,b对模m同余的充要条件是 a-b能被m整 阅读全文
posted @ 2016-04-18 23:27 超级学渣渣 阅读(3302) 评论(0) 推荐(0)

上一页 1 ··· 7 8 9 10 11 12 13 下一页