水题

题意： # W_H的方格纸，共有(w+1)_(H+1)个整点，现在将2个蜡烛放在2个不同的整点上。蜡烛不会被放在边界上。现在给出方格纸的尺寸和2个蜡烛的坐标，求一条线段将方格纸拆成2部分，而且这条线段不经过任何一个蜡烛且使得每一部分恰好有一个蜡烛。问线段的起点和终点。

题目链接 思路：注意xy-(x-2)*(y-2)=2x+2y-4，一定被2整除。因此siz为2的也是合法的。这个比较容易忘掉。 其他的判定条件都很好想。具体见代码； 代码实现 1#include <cstdio> 2#include <cstring> 3#include <iostream> 4#include <algorithm> 5#include <vector> 6#include <queue> 7#include <set> 8#include <map> 9#include <string> 10#include <cmath> 11#include <cstdlib> 12#include <ctime> 13#define fst first 14#define sec second 15#define lson l,m,rt<<1 16#define rson m+1,r,rt<<1|1 17#define ms(a,x) memset(a,x,sizeof(a)) 18typedef long long LL; 19#define pi pair < int ,int > 20#define MP make_pair 21using namespace std; 22const double eps = 1E-8; 23const int dx4[4]={1,0,0,-1}; 24const int dy4[4]={0,-1,1,0}; 25const int inf = 0x3f3f3f3f; 26int X,Y; 27bool ok( int a) 28{ 29 if (a==2) return true; 30 if (X%a==0&&(Y-2)%a==0) return true; 31 if (Y%a==0&&(X-2)%a==0) return true; 32 if (X%a==1&&Y%a==1) return true; 33 34 return false; 35} 36int main() 37{ 38 // freopen("in.txt","r",stdin); 39 40 while (~scanf("%d%d",&X,&Y)) 41 { 42 int n; 43 scanf("%d",&n); 44 while (n--) 45 { 46 int x; 47 scanf("%d",&x); 48 // cout<<"x:"<<x<<endl; 49 if (ok(x)) puts("YES"); 50 else puts("NO"); 51 } 52 } 53return 0; 54}

题目链接 題意：已知一个包含 nn 个元素的正整数集合 SS，设 f(S)f(S) 为集合 SS 中所有元素的异或(XOR)的结果。 如：S={1,2,3}, 则 f(S) = 0f(S)=0。 给出集合 SS，你需要计算 将所有 f(s)进行异或后的值， s⊆S.

题意：题意：给你一组三维空间中的点，每个点到其它点都有个距离，其中有个最小距离，如果这个最小距离小于10，就将对应的距离的点个数加１，最后输出距离为0,1,2…8,9的点的个数。（from 百度） 老实说，上面这题意也讲的不明不白，其实这题非常水，就是对每个点进行判断，找出和其他点最短的距离，在下标为该距离的数组上+1，最后输出数组下标0-9的数。 trick：其实最小距离大于9的就不用存放了，只要开个大小10的数组。（不会概括。。。抄的别人的）

http://codeforces.com/contest/611/problem/A 题意：两种查询，一种是 x of week,x为1.。7，对应输出2016年星期x有多少天。另一种为x of month ，对应输出2016年至少有x天的月份有多少天。 思路：直接搞。。。。竟然脑残被hack了。。。sad.

http://codeforces.com/contest/612/problem/C 题意：其实就是栈的基本操作。。水题。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2015年12月25日 星期五 22时58分50秒 4File Name :code/cf/edu4/C.cpp 5************************************************ */ 6 7#include <cstdio> 8#include <cstring> 9#include <iostream> 10#include <algorithm> 11#include <vector> 12#include <queue> 13#include <set> 14#include <map> 15#include <string> 16#include <cmath> 17#include <cstdlib> 18#include <ctime> 19#define fst first 20#define sec second 21#define lson l,m,rt<<1 22#define rson m+1,r,rt<<1|1 23#define ms(a,x) memset(a,x,sizeof(a)) 24typedef long long LL; 25#define pi pair < int ,int > 26#define MP make_pair 27 28using namespace std; 29const double eps = 1E-8; 30const int dx4[4]={1,0,0,-1}; 31const int dy4[4]={0,-1,1,0}; 32const int inf = 0x3f3f3f3f; 33const int N=1E6+7; 34int len; 35char st[N]; 36int cost = 0 ; 37 38char a[N]; 39int n ; 40 41 42int which(char ch) 43{ 44 if (ch=='<'||ch=='{'||ch=='('||ch=='[') return 1; 45 return 2; 46} 47int kin(char ch) 48{ 49 if (ch=='{'||ch=='}') return 1; 50 if (ch=='['||ch==']') return 2; 51 if (ch=='<'||ch=='>') return 3; 52 if (ch=='('||ch==')') return 4; 53} 54bool ok(char x,char y) 55{ 56 int res = 0 ; 57 if(x=='<'||x=='{'||x=='['||x=='(') res++; 58 if (y=='>'||y=='}'||y==']'||y==')') res++; 59 if (res==2) 60 { 61 if (kin(x)!=kin(y)) cost++; 62// cout<<"x:"<<x<<" y:"<<y<<endl; 63 return true; 64 } 65 return false; 66} 67 68int main() 69{ 70 #ifndef ONLINE_JUDGE 71 freopen("code/in.txt","r",stdin); 72 #endif 73 cin>>st; 74 len = strlen(st); 75 int head = -1; 76 int ans = 0 ; 77 78 for ( int i = 0 ; i < len ; i++) 79 { 80 if (head==-1) 81 { 82 head++; 83 a[head] = st[i]; 84 if (which(st[i])==2) 85 { 86 puts("Impossible"); 87 return 0; 88 } 89 90 continue; 91 } 92 if (ok(a[head],st[i])) 93 { 94 head--; 95 } 96 else 97 { 98 head++; 99 a[head] = st[i]; 100 } 101 102// cout<<"head:"<<head<<endl; 103 } 104// cout<<"head:"<<head<<endl; 105 if (head!=-1) 106 { 107 puts("Impossible"); 108 } 109 else 110 { 111 cout<<cost<<endl; 112 } 113 114 #ifndef ONLINE_JUDGE 115 fclose(stdin); 116 #endif 117 return 0; 118}

http://codeforces.com/contest/612/problem/B 水。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2015年12月25日 星期五 22时58分38秒 4File Name :code/cf/edu4/B.cpp 5************************************************ */ 6 7#include <cstdio> 8#include <cstring> 9#include <iostream> 10#include <algorithm> 11#include <vector> 12#include <queue> 13#include <set> 14#include <map> 15#include <string> 16#include <cmath> 17#include <cstdlib> 18#include <ctime> 19#define fst first 20#define sec second 21#define lson l,m,rt<<1 22#define rson m+1,r,rt<<1|1 23#define ms(a,x) memset(a,x,sizeof(a)) 24typedef long long LL; 25#define pi pair < int ,int > 26#define MP make_pair 27 28using namespace std; 29const double eps = 1E-8; 30const int dx4[4]={1,0,0,-1}; 31const int dy4[4]={0,-1,1,0}; 32const int inf = 0x3f3f3f3f; 33const int N=2E5+7; 34int n; 35struct node 36{ 37 int val; 38 int id; 39 40 bool operator <(node b)const 41 { 42 return val<b.val; 43 } 44}q[N]; 45int main() 46{ 47 #ifndef ONLINE_JUDGE 48 freopen("code/in.txt","r",stdin); 49 #endif 50 cin>>n; 51 for ( int i = 1 ; i <= n ; i++ ) 52 { 53 scanf("%d",&q[i].val); 54 q[i].id = i; 55 } 56 sort(q+1,q+n+1); 57 LL ans = 0 ; 58 q[0].id = 0 ; 59 for ( int i = 1 ; i <= n-1 ; i++) 60 { 61 ans += LL (abs(q[i+1].id-q[i].id)); 62 } 63 cout<<ans<<endl; 64 65 #ifndef ONLINE_JUDGE 66 fclose(stdin); 67 #endif 68 return 0; 69}

http://codeforces.com/contest/612/problem/A 水题…直接枚举就好。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2015年12月25日 星期五 22时58分26秒 4File Name :code/cf/edu4/A.cpp 5************************************************ */ 6 7#include <cstdio> 8#include <cstring> 9#include <iostream> 10#include <algorithm> 11#include <vector> 12#include <queue> 13#include <set> 14#include <map> 15#include <string> 16#include <cmath> 17#include <cstdlib> 18#include <ctime> 19#define fst first 20#define sec second 21#define lson l,m,rt<<1 22#define rson m+1,r,rt<<1|1 23#define ms(a,x) memset(a,x,sizeof(a)) 24typedef long long LL; 25#define pi pair < int ,int > 26#define MP make_pair 27 28using namespace std; 29const double eps = 1E-8; 30const int dx4[4]={1,0,0,-1}; 31const int dy4[4]={0,-1,1,0}; 32const int inf = 0x3f3f3f3f; 33const int N=105; 34int n,p,q; 35char st[N]; 36bool v[10005]; 37int main() 38{ 39 #ifndef ONLINE_JUDGE 40 freopen("code/in.txt","r",stdin); 41 #endif 42 cin>>n>>p>>q; 43 cin>>st; 44 ms(v,false); 45 int b,c; 46 for ( int i = 0 ; i*p<=105 ; i++ ) 47 { 48 for ( int j = 0 ; j*q <= 105 ; j++) 49 { 50 v[i*p+j*q] = true; 51 if (i*p+j*q==n) 52 { 53 b = i; 54 c = j; 55 } 56 } 57 } 58// cout<<"b:"<<b<<endl; 59// cout<<"c:"<<c<<endl; 60 if (!v[n]) 61 { 62 puts("-1"); 63 } 64 else 65 { 66 int cnt = 0 ; 67 printf("%d\n",b+c); 68 for ( int i = 1 ; i <= b ; i++) 69 { 70 for ( int i = 0 ; i < p ; i++) 71 printf("%c",st[cnt]),cnt++; 72 printf("\n"); 73 } 74 for ( int i =1 ; i <= c ; i++) 75 { 76 for ( int i = 0 ; i < q ; i++) 77 printf("%c",st[cnt]),cnt++; 78 printf("\n"); 79 } 80 } 81 82 83 84 #ifndef ONLINE_JUDGE 85 fclose(stdin); 86 #endif 87 return 0; 88}

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4704 题目大意是说，定义一个数的lucky number是距离i最远的j且满足（a[i]<a[j] i<j）。 问对所有数最大的lucky number是什么。 不必线段树。

I - Lottery **Time Limit:**2000MS **Memory Limit:**524288KB 64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 589I Description Today Berland holds a lottery with a prize – a huge sum of money! There are k persons, who attend the lottery. Each of them will receive a unique integer from 1 to k. The organizers bought n balls to organize the lottery, each of them is painted some color, the colors are numbered from 1 to k. A ball of color c corresponds to the participant with the same number. The organizers will randomly choose one ball – and the winner will be the person whose color will be chosen!

A. Duff and Meat time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Duff is addicted to meat! Malek wants to keep her happy for n days. In order to be happy in i-th day, she needs to eat exactly a__i kilograms of meat. There is a big shop uptown and Malek wants to buy meat for her from there. In i-th day, they sell meat for p__i dollars per kilogram. Malek knows all numbers _a_1, …, a__n and _p_1, …, p__n. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.

A. Vasya the Hipster time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Vasya the Hipster decided to count how many socks he had. It turned out that he had a red socks and b blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn’t want to wash them.

In ``Hangman Judge,’’ you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follows: 1. The contestant tries to solve to puzzle by guessing one letter at a time. 2. Every time a guess is correct, all the characters in the word that match the guess will be ``turned over.'' For example, if your guess is ``o'' and the word is ``book'', then both ``o''s in the solution will be counted as ``solved.'' 3. Every time a wrong guess is made, a stroke will be added to the drawing of a hangman, which needs 7 strokes to complete. Each unique wrong guess only counts against the contestant once. ______ | | | O | /| | | | / __|_ | |______ |_________| 4. If the drawing of the hangman is completed before the contestant has successfully guessed all the characters of the word, the contestant loses. 5. If the contestant has guessed all the characters of the word before the drawing is complete, the contestant wins the game. 6. If the contestant does not guess enough letters to either win or lose, the contestant chickens out. Your task as the ``Hangman Judge’’ is to determine, for each game, whether the contestant wins, loses, or fails to finish a game.

由于顺序是可以改变的． 所以考虑是否可以映射．只要存在字母对应出现的次数都相同．那么就可以通过映射得到． 具体是开一个数组记录每个字母出现的次数… 然后sort

应该３分钟过的题。。。 结果花了８分钟。。。ssssad 代码实现 1/************************************************************************* 2 > File Name: code/cf/#317/A.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月23日 星期日 00时29分47秒 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#include<cctype> 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define ms(a,x) memset(a,x,sizeof(a)) 25#define lr dying111qqz 26using namespace std; 27#define For(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef double DB; 30const int inf = 0x3f3f3f3f; 31const int N=1E5+7; 32int na,nb,k,m; 33int a[N],b[N]; 34int main() 35{ 36 #ifndef ONLINE_JUDGE 37 freopen("in.txt","r",stdin); 38 #endif 39 cin>>na>>nb; 40 cin>>k>>m; 41 for ( int i = 1 ; i <= na ; i++){ 42 scanf("%d",&a[i]); 43 } 44 for ( int i = 1 ; i<= nb ; i++){ 45 scanf("%d",&b[i]); 46 } 47 if (a[k]<b[nb-m+1]){ 48 puts("YES"); 49 } 50 else 51 { 52 puts("NO"); 53 } 54 55 56 #ifndef ONLINE_JUDGE 57 fclose(stdin); 58 #endif 59 return 0; 60}