cstdio

[uva816]AbbottsRevenge Abbott的复仇(经典迷宫BFS)

这题思路就普通的BFS加上一个维度朝向，主要是要注意输入，输出，以及细节的处理   #include<cstdio> #include<cstring> #include
·2015-11-12 17:59

HDU 3746 Cyclic Nacklace KMP

pid=3746 KMP算法—— AC代码：   #include <iostream> #include <cstdio> #include
·2015-11-12 17:14

CF 191 div2

  #include <iostream> #include <cstdio> #include <algorithm> #include <
·2015-11-12 17:10

HDU 1358 Period KMP

pid=1358 求周期问题，简单KMP—— AC代码：   #include <iostream> #include <cstdio> #include
·2015-11-12 17:06

10716 - Evil Straw Warts Live

描述：把所给的字符串变换成回文，并统计其变换次数，如果不能，则输出Impossible #include <cstdio> #include <cstdlib> #include
·2015-11-12 17:41

POJ-2001-Shortest Prefixes

id=2001 找出能唯一标示一个字符串的最短前缀，如果找不出，就输出该字符串 用字典树即可 #include<iostream> #include<cstdio> #
·2015-11-12 17:26

poj 3269 Building A New Barn

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm&
·2015-11-12 16:12

UVA 11584 Partitioning by Palindromes 动态规划 入门

#include<iostream> #include<cstdio> #include<cstring
·2015-11-12 16:37

波浪子序列 Wavio Sequence UVA10534 动态规划 最长上升子序列变形

#include<iostream> #include<cstdio> #includ
·2015-11-12 16:36

cake 简单思维题

  1 #include <iostream> 2 #include <cstdio> 3 #include <cstring>
·2015-11-12 16:00

Undraw the Trees 字符串数组递归画树

1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #inclu
·2015-11-12 16:59

[DP] LCS小结

  DP、O(n^2)解法： #include <iostream> #include <cstdio> #include <cstring>
·2015-11-12 16:41

HDU 1426 Sudoku Killer

#include<cstdio> #include<cstring> #include<string> #include<queue> #include&
·2015-11-12 16:20

hdu 3756 Dome of Circus(三分)

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:54

hdu 2289 cup(二分搜索)

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:53

hdu 3732 Ahui Writes Word(多重背包)

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:53

hdu 3033 I love sneakers!（01背包变种）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:52

hdu 3496 Watch The Movie(二维费用背包)

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:51

hdu 1712 ACboy needs your help(分组背包)

分组背包一水~ View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio>
·2015-11-12 16:51

hdu 1133 Buy the Ticket（大数运算）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:50

hdu 1130 How Many Trees?（卡特兰数）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:48

hdu 1063 Exponentiation（高精度幂）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #
·2015-11-12 16:48

Codeforces Round #149 (Div. 2)

做法：bfs+map View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio>
·2015-11-12 16:46

UVA 11584 Partitioning by Palindromes

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:45

hdu 2955 Robberies(01背包)

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:42

hdu 1175 连连看（dfs+剪枝）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:42

hdu 1069 Monkey and Banana

+ h[i] ; (j<i) View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio
·2015-11-12 16:41

hdu 4031 Attack(树状数组)

ans = 总的攻击次数 - 防御次数 View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio
·2015-11-12 16:40

poj 1195 Mobile phones（二维树状数组）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:39

poj 1753 Flip Game（枚举）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:38

hdu 4417 Super Mario(树状数组+离线处理)

171MS View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio>
·2015-11-12 16:37

hdu 4034 Graph（floyd）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:37

poj 1273 Drainage Ditches（网络流基础）

Ford_Fullkerson 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio>
·2015-11-12 16:35

zoj 2671 Cryptography（矩阵+线段树）

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:34

SRM 556 DIV2

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:33

hdu 4027 Can you answer these queries?

View Code 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include
·2015-11-12 16:32

poj 3368 Frequent values

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #
·2015-11-12 16:30

hdu 1068 Girls and Boys

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:29

poj 2239 Selecting Courses

View Code 1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include
·2015-11-12 16:28

poj 3348 Cows

三角形面积等于叉积的一半~ View Code 1 /* 2 Coder:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio
·2015-11-12 16:27

poj 1113 Wall

凸包周长+圆周长 View Code 1 /* 2 Coder:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio>
·2015-11-12 16:27

HDU 3415 Max Sum of Max-K-sub-sequence

#include <cstdio> using namespace std; int a[100005],s[200010],num[200010]; int T,n,m,st
·2015-11-12 16:38

HDU 1575 Tr A

基础矩阵乘法…… #include <cstdio> #include <iostream> #define rep(i,n) for(int i=0;i<n;i++
·2015-11-12 16:36

HDU 1851 A Simple Game

#include <cstdio> int main(){ int T,i,n,SG,m,l; scanf("%d",&T); while
·2015-11-12 16:35

HDU 2147 kiki's game

题解：画图可得当横纵坐标均为奇数时为必败态……                #include <cstdio&
·2015-11-12 16:34

HDU 2188 悼念512汶川大地震遇难同胞——选拔志愿者

题解：最简单的巴什博弈，取石子游戏…… #include <cstdio> int main(){ int T,n,m; scanf("%d",&
·2015-11-12 16:34

HDU 1850 Being a Good Boy in Spring Festival

#include <cstdio> using namespace std; int SG[105]; int main(){ int n,sum
·2015-11-12 16:33

HDU 1848 Fibonacci again and again

#include <cstdio> #include <cstring> using namespace std; int fbi[30]; int SG[
·2015-11-12 16:32

HDU 1849 Rabbit and Grass

#include <cstdio> int main(){ int SG,n,a; while(scanf("%d",&n),n){
·2015-11-12 16:32

HDU 1846 Brave Game

题解：巴什博弈，考虑Ｐ点为０，则Ｎ点在１至ｍ之间，则k(m+1)＋ｒ为Ｎ点，所以只要满足ｎ％（ｍ＋１）＝＝０则先手必败，反之则先手必胜…… #include <cstdio> int main
·2015-11-12 16:31
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