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2015

bc #43(hdu 5265) pog loves szh II (单调性优化)

·624 words·2 mins
pog loves szh II # **Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2115 Accepted Submission(s): 609 ** Problem Description Pog and Szh are playing games.There is a sequence with n numbers, Pog will choose a number A from the sequence. Szh will choose an another number named B from the rest in the sequence. Then the score will be (A+B) mod p.They hope to get the largest score.And what is the largest score?

sgu 463 - Walking around Berhattan

·625 words·2 mins
简单模拟,n,m貌似给反了(两个地方给的不一致 ) 害我wa了两发 代码实现 1 2 /************************************************************************* 3 > File Name: code/2015summer/#5/K.cpp 4 > Author: 111qqz 5 > Email: rkz2013@126.com 6 > Created Time: 2015年07月30日 星期四 14时00分56秒 7 ************************************************************************/ 8 9 #include<iostream> 10 #include<iomanip> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cmath> 14 #include<cstring> 15 #include<string> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #define y0 abc111qqz 22 #define y1 hust111qqz 23 #define yn hez111qqz 24 #define j1 cute111qqz 25 #define tm crazy111qqz 26 #define lr dying111qqz 27 using namespace std; 28 #define REP(i, n) for (int i=0;i<int(n);++i) 29 typedef long long LL; 30 typedef unsigned long long ULL; 31 const int inf = 0x7fffffff; 32 const int N=1e2+5; 33 int b[N][N]; 34 int n,m; 35 char cmd[505]; 36 bool vis[N][N]; 37 int nx,ny; 38 int dx[4]={-1,0,1,0}; 39 int dy[4]={0,1,-0,-1}; 40 char ch[N][N]; 41 int main() 42 { 43 cin>>n>>m; 44 nx = 0; 45 ny = 0; 46 for ( int i = 0 ; i < n ; i++) 47 cin>>ch[i]; 48 for ( int i = 0 ; i <n ; i++ ) 49 { 50 for ( int j = 0 ; j < m ; j++ ) 51 { 52 b[i+1][j+1]=(int)(ch[i][j]-'0'); 53 } 54 } 55 int ans = 0; 56 memset(vis,false,sizeof(vis)); 57 int dir = 1; 58 cin>>cmd; 59 int len = strlen(cmd); 60 for ( int i = 0 ; i < len ; i ++ ) 61 { 62 // cout<<"ans:"<<ans<<endl; 63 // cout<<"nx:"<<nx<<" ny:"<<ny<<endl; 64 if (cmd[i]=='L') 65 { 66 dir = (dir+3)%4; 67 } 68 if (cmd[i]=='R') 69 { 70 dir = (dir+1)%4; 71 } 72 if (cmd[i]=='M') 73 { 74 if (dir==0) 75 { 76 if (vis[nx][ny]) 77 { 78 ans = ans + b[nx][ny]/2; 79 } 80 else 81 { 82 ans = ans + b[nx][ny]; 83 } 84 if (vis[nx][ny+1]) 85 { 86 ans = ans + b[nx][ny+1]/2; 87 } 88 else 89 { 90 ans = ans + b[nx][ny+1]; 91 } 92 vis[nx][ny]=true; 93 vis[nx][ny+1]=true; 94 nx = nx +dx[dir]; 95 ny = ny +dy[dir]; 96 } 97 if (dir==2) 98 { 99 nx = nx + dx[dir]; 100 ny = ny + dy[dir]; 101 if (vis[nx][ny]) 102 { 103 ans = ans + b[nx][ny]/2; 104 } 105 else 106 { 107 ans = ans + b[nx][ny]; 108 } 109 if (vis[nx][ny+1]) 110 { 111 ans = ans + b[nx][ny+1]/2; 112 } 113 else 114 { 115 ans = ans + b[nx][ny+1]; 116 } 117 vis[nx][ny]=true; 118 vis[nx][ny+1]=true; 119 } 120 if (dir==1) 121 { 122 nx = nx + dx[dir]; 123 ny = ny + dy[dir]; 124 if (vis[nx][ny]) 125 { 126 ans = ans + b[nx][ny]/2; 127 } 128 else 129 { 130 ans = ans + b[nx][ny]; 131 } 132 if (vis[nx+1][ny]) 133 { 134 ans = ans + b[nx+1][ny]/2; 135 } 136 else 137 { 138 ans = ans + b[nx+1][ny]; 139 } 140 vis[nx][ny]=true; 141 vis[nx+1][ny]=true; 142 143 } 144 if (dir==3) 145 { 146 147 if (vis[nx][ny]) 148 { 149 ans = ans + b[nx][ny]/2; 150 } 151 else 152 { 153 ans = ans + b[nx][ny]; 154 } 155 if (vis[nx+1][ny]) 156 { 157 ans = ans + b[nx+1][ny]/2; 158 } 159 else 160 { 161 ans = ans + b[nx+1][ny]; 162 } 163 vis[nx][ny]=true; 164 vis[nx+1][ny]=true; 165 nx = nx +dx[dir]; 166 ny = ny +dy[dir]; 167 } 168 } 169 170 } 171 cout<<ans<<endl; 172 173 return 0; 174 }

SGU 456 Annuity Payment Scheme

·252 words·1 min
水题,推个公式出来,注意精度…一遍A 代码实现 1 2 /************************************************************************* 3 > File Name: code/2015summer/#5/D.cpp 4 > Author: 111qqz 5 > Email: rkz2013@126.com 6 > Created Time: 2015年07月30日 星期四 13时17分26秒 7 ************************************************************************/ 8 9 #include<iostream> 10 #include<iomanip> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cmath> 14 #include<cstring> 15 #include<string> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #define y0 abc111qqz 22 #define y1 hust111qqz 23 #define yn hez111qqz 24 #define j1 cute111qqz 25 #define tm crazy111qqz 26 #define lr dying111qqz 27 using namespace std; 28 #define REP(i, n) for (int i=0;i<int(n);++i) 29 typedef long long LL; 30 typedef unsigned long long ULL; 31 const int inf = 0x7fffffff; 32 int s,m,p; 33 double ans; 34 35 double cal(double x,int n) 36 { 37 double res = 1.0; 38 for ( int i = 1 ; i <= n ; i++ ) 39 { 40 res = res * x; 41 } 42 // cout<<"res:"<<res<<endl; 43 return res; 44 } 45 int main() 46 { 47 cin>>s>>m>>p; 48 double sum = 0; 49 double per = p*1.0/100+1; 50 for ( int i = 1 ; i <= m; i++ ) 51 { 52 sum=sum+1.0/cal(per,i); 53 // cout<<"sum:"<<sum<<endl; 54 } 55 // cout<<sum<<endl; 56 cout<<fixed<<setprecision(5)<<s*1.0/sum<<endl; 57 58 return 0; 59 }

SPOJ AMR10F Cookies Piles

·162 words·1 min
AMR10F - Cookies Piles # 水. 1 2 3 /************************************************************************* 4 > File Name: code/2015summer/#4/F.cpp 5 > Author: 111qqz 6 > Email: rkz2013@126.com 7 > Created Time: 2015年07月29日 星期三 21时47分23秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<iomanip> 12 #include<cstdio> 13 #include<algorithm> 14 #include<cmath> 15 #include<cstring> 16 #include<string> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 #define y0 abc111qqz 23 #define y1 hust111qqz 24 #define yn hez111qqz 25 #define j1 cute111qqz 26 #define tm crazy111qqz 27 #define lr dying111qqz 28 using namespace std; 29 #define REP(i, n) for (int i=0;i<int(n);++i) 30 typedef long long LL; 31 typedef unsigned long long ULL; 32 const int inf = 0x7fffffff; 33 int main() 34 { 35 int T; 36 int n,a,d; 37 cin>>T; 38 while (T--) 39 { 40 scanf("%d %d %d",&n,&a,&d); 41 cout<<n*a+n*(n-1)/2*d<<endl; 42 } 43 44 return 0; 45 }

poj 2823 Sliding Window (单调队列)

·1773 words·4 mins
Sliding Window 看这个问题:An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position.Your task is to determine the maximum and minimum values in the sliding window at each position. 也就是有一个数列a,要求你求数列b和c,b[i]是a[i]…a[i+w-1]中的最小值,c[i]是最大值。如果a是1,3,-1,-3,5,3,6,7,则b为-1,-3,-3,-3,3,3,c为3,3,5,5,6,7。

codeforces 442C. Artem and Array

·492 words·1 min
C. Artem and Array time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Artem has an array of n positive integers. Artem decided to play with it. The game consists of n moves. Each move goes like this. Artem chooses some element of the array and removes it. For that, he gets min(a, b) points, where a and b are numbers that were adjacent with the removed number. If the number doesn’t have an adjacent number to the left or right, Artem doesn’t get any points.

cf 442B Andrey and Problem

·712 words·2 mins
B. Andrey and Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends – the probability that this friend will come up with a problem if Andrey asks him.

cf 443B Kolya and Tandem Repeat

·445 words·1 min
B. Kolya and Tandem Repeat time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Kolya got string s for his birthday, the string consists of small English letters. He immediately added k more characters to the right of the string. Then Borya came and said that the new string contained a tandem repeat of length l as a substring. How large could l be? See notes for definition of a tandem repeat.

(BC 一周年) hdu 5312 Sequence

·767 words·2 mins
比赛的时候没做出来.这道题需要用到的一个重要的性质是,任意一个自然数可以表示成至多三个三角形数(1,3,6,10,15…..)的和(orz高斯)然后也有推广到任意自然数可以表示成k个k角形数的和的结论(费马提出了猜想,柯西给了证明)然后官方题解说的比较好:

三角形数_百度百科

·876 words·2 mins
它有一定的规律性,排列如下(构成图),像上面的1、3、6、10、15等等这些能够表示成三角形的形状的总数量的数,叫做三角形数。 一定数目的点或圆在等距离的排列下可以形成一个等边三角形,这样的数被称为三角形数。比如10个点可以组成一个等边三角形,因此10是一个三角形数:

(BC 一周年)hdu 5311 Hidden String

·767 words·2 mins
http://acm.hdu.edu.cn/showproblem.php?pid=5311 题意:问能否从一个给定的字符串中拿出三个不相交的字串(原串可以有剩余),组成字符串“anniversary” 思路:暴力。 比赛的时候没做出来,sad 我发现我有一个问题,就是不敢跑暴力 有不少题其实正解就是暴力 或者有的题,暴力不是标解,但是绝对可A,可我就不敢写… 就觉得不会是这样.. 说到底还是不自信吧…

(BC 一周年)hdu 5310 Souvenir

·186 words·1 min
http://acm.hdu.edu.cn/showproblem.php?pid=5310 水。 不要用cin. 代码实现 1 2 /************************************************************************* 3 > File Name: code/bc/#ann/1001.cpp 4 > Author: 111qqz 5 > Email: rkz2013@126.com 6 > Created Time: 2015年07月25日 星期六 18时54分24秒 7 ************************************************************************/ 8 9 #include<iostream> 10 #include<iomanip> 11 #include<cstdio> 12 #include<algorithm> 13 #include<cmath> 14 #include<cstring> 15 #include<string> 16 #include<map> 17 #include<set> 18 #include<queue> 19 #include<vector> 20 #include<stack> 21 #define y0 abc111qqz 22 #define y1 hust111qqz 23 #define yn hez111qqz 24 #define j1 cute111qqz 25 #define tm crazy111qqz 26 #define lr dying111qqz 27 using namespace std; 28 #define REP(i, n) for (int i=0;i<int(n);++i) 29 typedef long long LL; 30 typedef unsigned long long ULL; 31 int n,m,p,q; 32 int main() 33 { 34 int T; 35 cin>>T; 36 int ans = 0; 37 while (T--) 38 { 39 // scanf("%d %d %d %d",&n,&m,&p,&q); 40 scanf("%d %d %d %d",&n,&m,&p,&q); 41 ans = n*p; 42 ans = min(ans,n/m*q+n%m*p); 43 ans = min(ans,((n-1)/m+1)*q); 44 printf("%d\n",ans); 45 } 46 47 return 0; 48 }

uva 12442 . Forwarding Emails

·579 words·2 mins
“… so forward this to ten other people, to prove that you believe the emperor has 题意是说发短信,每个人只会给一个人发,问从哪个人开始发,能传到的人最多 思路是每个人开始做一遍dfs… 毫无意外的TLE了 一个容易想到的剪枝是,如果在第i次之前的路径上的点,在之后以它作为起点遍历一定不优.

I - Fire Game (两个点开始的bfs)

·1278 words·3 mins
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83084#problem/I I - Fire Game **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)

poj 3414 pots (bfs+路径记录)

·496 words·1 min
好爽,一遍ac 代码实现 1 2 3 /************************************************************************* 4 > File Name: code/2015summer/searching/H.cpp 5 > Author: 111qqz 6 > Email: rkz2013@126.com 7 > Created Time: 2015年07月27日 星期一 09时11分28秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<iomanip> 12 #include<cstdio> 13 #include<algorithm> 14 #include<cmath> 15 #include<cstring> 16 #include<string> 17 #include<map> 18 #include<set> 19 #include<queue> 20 #include<vector> 21 #include<stack> 22 #define y0 abc111qqz 23 #define y1 hust111qqz 24 #define yn hez111qqz 25 #define j1 cute111qqz 26 #define tm crazy111qqz 27 #define lr dying111qqz 28 using namespace std; 29 #define REP(i, n) for (int i=0;i<int(n);++i) 30 typedef long long LL; 31 typedef unsigned long long ULL; 32 const int N=1E2+5; 33 int A,B,C; 34 int d[N][N]; 35 bool flag; 36 struct node 37 { 38 int d,opt,par,prea,preb; 39 }q[N][N]; 40 41 void print(int x,int y) 42 { 43 // cout<<"x:"<<x<<"y:"<<y<<endl; 44 if (q[x][y].prea!=-1&&q[x][y].preb!=-1) 45 { 46 // cout<<"who is 111qqz"<<endl; 47 print(q[x][y].prea,q[x][y].preb); 48 if (q[x][y].opt==1){ 49 printf("FILL(%d)\n",q[x][y].par); 50 } 51 if (q[x][y].opt==2) 52 { 53 printf("DROP(%d)\n",q[x][y].par); 54 } 55 if (q[x][y].opt==3) 56 { 57 printf("POUR(%d,%d)\n",q[x][y].par,3-q[x][y].par); 58 } 59 } 60 61 } 62 void bfs() 63 { 64 memset(q,-1,sizeof(q)); 65 queue<int>a; 66 queue<int>b; 67 a.push(0); 68 b.push(0); 69 q[0][0].d=0; 70 while (!a.empty()&&!b.empty()) 71 { 72 int av = a.front();a.pop(); 73 int bv = b.front();b.pop(); 74 // cout<<"av:"<<av<<"bv:"<<bv<<endl; 75 if (av==C||bv==C) 76 { 77 flag = true; 78 //cout<<"yeah~~~~~~~~~~~~~~~"<<endl; 79 cout<<q[av][bv].d<<endl; 80 // cout<<"prea:"<<q[av][bv].prea<<"preb:"<<q[av][bv].preb<<endl; 81 print(av,bv); 82 return; 83 } 84 if (av<A&&q[A][bv].d==-1) 85 { 86 q[A][bv].d=q[av][bv].d+1; 87 q[A][bv].opt=1; 88 q[A][bv].par=1; 89 q[A][bv].prea=av; 90 q[A][bv].preb=bv; 91 a.push(A); 92 b.push(bv); 93 } 94 if (av>0&&q[0][bv].d==-1) 95 { 96 q[0][bv].d=q[av][bv].d+1; 97 q[0][bv].opt=2; 98 q[0][bv].par=1; 99 q[0][bv].prea=av; 100 q[0][bv].preb=bv; 101 a.push(0); 102 b.push(bv); 103 104 } 105 if (bv<B&&q[av][B].d==-1) 106 { 107 q[av][B].d=q[av][bv].d+1; 108 q[av][B].opt=1; 109 q[av][B].par=2; 110 q[av][B].prea = av; 111 q[av][B].preb = bv; 112 a.push(av); 113 b.push(B); 114 115 } 116 if (bv>0&&q[av][0].d==-1) 117 { 118 q[av][0].d=q[av][bv].d+1; 119 q[av][0].opt=2; 120 q[av][0].par=2; 121 q[av][0].prea=av; 122 q[av][0].preb=bv; 123 a.push(av); 124 b.push(0); 125 } 126 127 if (av+bv<=B&&q[0][av+bv].d==-1) 128 { 129 q[0][av+bv].d=q[av][bv].d+1; 130 q[0][av+bv].opt=3; 131 q[0][av+bv].par=1; 132 q[0][av+bv].prea=av; 133 q[0][av+bv].preb=bv; 134 a.push(0); 135 b.push(av+bv); 136 } 137 if (av+bv>B&&q[av-(B-bv)][B].d==-1) //把1往2里倒入的两种情况 138 { 139 140 int tmp = av-(B-bv); 141 q[tmp][B].d=q[av][bv].d+1; 142 q[tmp][B].opt=3; 143 q[tmp][B].par=1; 144 q[tmp][B].prea=av; 145 q[tmp][B].preb=bv; 146 a.push(tmp); 147 b.push(B); 148 } 149 150 if (bv+av<=A&&q[av+bv][0].d==-1) 151 { 152 q[av+bv][0].d=q[av][bv].d+1; 153 q[av+bv][0].opt=3; 154 q[av+bv][0].par=2; 155 q[av+bv][0].prea=av; 156 q[av+bv][0].preb=bv; 157 a.push(av+bv); 158 b.push(0); 159 } 160 if (bv+av>A&&q[A][bv-(A-av)].d==-1) 161 { 162 int tmp = bv-(A-av); 163 q[A][tmp].d=q[av][bv].d+1; 164 q[A][tmp].opt=3; 165 q[A][tmp].par=2; 166 q[A][tmp].prea=av; 167 q[A][tmp].preb=bv; 168 a.push(A); 169 b.push(tmp); 170 } 171 } 172 } 173 int main() 174 { 175 176 flag = false; 177 cin>>A>>B>>C; 178 bfs(); 179 if (!flag) 180 { 181 cout<<"impossible"<<endl; 182 } 183 184 185 return 0; 186 }

hdoj 1495 非常可乐(bfs)

·1015 words·3 mins
非常可乐 # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194 Accepted Submission(s): 2865 ** Problem Description 大家一定觉的运动以后喝可乐是一件很惬意的事情,但是seeyou却不这么认为。因为每次当seeyou买了可乐以后,阿牛就要求和seeyou一起分享这一瓶可乐,而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子,它们的容量分别是N 毫升和M 毫升 可乐的体积为S (S<101)毫升 (正好装满一瓶) ,它们三个之间可以相互倒可乐 (都是没有刻度的,且 S==N+M,101>S>0,N>0,M>0) 。聪明的ACMER你们说他们能平分吗?如果能请输出倒可乐的最少的次数,如果不能输出"NO"。

hdoj 1241 Oil Deposits (dfs)

·628 words·2 mins
Oil Deposits # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17683 Accepted Submission(s): 10172 ** Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

hdoj 2612 find a way (两次bfs)

·801 words·2 mins
Find a way # Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6221 Accepted Submission(s): 2070 Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.