better_space的博客

#include<stdio.h>__int64 quickpow(int m,int n,int s) { __int64 ans=1,base=m; while(n) { if(n&1) { ans=ans*base%s; } base=base*base%s; n>>=1; }

Find them, Catch themTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 42064 Accepted: 12936DescriptionThe police office in Tadu City decides to say end

Apple TreeTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 26622 Accepted: 7900DescriptionThere is an apple tree outside of kaka's house. Every autumn, a

LabyrinthTime Limit: 2000MS Memory Limit: 32768KTotal Submissions: 4333 Accepted: 1640DescriptionThe northern part of the Pyramid contains a very large and comp

Longest Ordered SubsequenceTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 47040 Accepted: 20904DescriptionA numeric sequence of ai is ordered if a1 a2

DivisibilityTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11568 Accepted: 4155DescriptionConsider an arbitrary sequence of integers. One can place + o

Largest Rectangle in a HistogramTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19452 Accepted: 6267DescriptionA histogram is a polygon composed of a

Bad Hair DayTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 17576 Accepted: 5926DescriptionSome of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a

Difference Is BeautifulTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 2247 Accepted: 703DescriptionMr. Flower's business is growing much faster than

Balanced LineupTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 47436 Accepted: 22281Case Time Limit: 2000MSDescriptionFor the daily milking, Far

Balanced LineupTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 47142 Accepted: 22121Case Time Limit: 2000MSDescriptionFor the daily milking, Far

OulipoTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35904 Accepted: 14492DescriptionThe French author Georges Perec (1936–1982) once wrote a book, La

Bungee JumpingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1365 Accepted: 604DescriptionOnce again, James Bond is fleeing from some evil people who

题意:从’@‘开始找上下左右相邻的’.’的个数。#include<stdio.h>const int N = 22;char c[N][N];int m,n,ans;int px[4]= {1,-1,0,0};int py[4]= {0,0,1,-1};void dfs(int x,int y) { c[x][y]='#'; ans++; for(int l=0;

用BFS简单些，无非就是吧二维数组变成三维数组，过程和其他求是否连通的题目大同小异。 坐标变化是6. 不过这题我写的时候直接在标记出压缩成一行判断，老是得不到正确结果，再看了大神的代码后，增加了check（）这一判断函数，就奇迹般的ac了，所以以后对于复杂的判断最好写一个判断函数。 另外我发现大神代码的另一个优点，能减少输入后便利取S和E的循环次数，这种小的优化很重要，值得借鉴。（下方代码中“

搜索里应用型的题目。 记得考虑临界情况哦。 BFS代码：#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>using namespace std;const int N = 100000;bool vis[N+100];int n,m;struct note { int t;

求一个坐标区域内，在给出的几点坐标中上下左右相邻的个数的最大值。#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int N = 105;int map[N][N];int vis[N][N];int n,m,k,ans;int dir[4][2] = {1, 0, -1,

这题虽是二分的题，但感觉就是别扭，很难想清楚怎么设置判断。我也是看了别人的判断过程及判断依据后才成功ac。值得收藏的二分题。 判断的过程我尽量解释一下，有不清楚的地方请参阅网址： http://www.cnblogs.com/jingqi814/p/3581553.html这样讲应该能理解的： 1.首先去掉M个石头后两个石头之间会有最小的距离，本题让求这个最小距离的最大值。 2.我们二分是在

暂且只用DFS做着，用BFS我老wa“哇。。桑心” DFS：#include<stdio.h>#include<string.h>const int N = 105;int a[N][N],b[N][N];char c[N][N];int px[8]= {0,0,1,1,1,-1,-1,-1};int py[8]= {-1,1,-1,0,1,-1,0,1};int m,n;void