cf #309a A - Kyoya and Photobooks

http://codeforces.com/contest/556/problem/C 果然一晚上不睡觉会导致读错题么… 需要注意的是 如果有一个是 1 2 4 6 那么 1,2是不必拆开的…. 然后我们发现，只有以1为开始且连续的套娃不必拆开….

http://codeforces.com/problemset/problem/548/A 水题。分割成K个，每个串判断是否回文，如果都是就yes，否则no 需要注意的是，可能不能正好分成长度相同的K个，这个时候也要No 代码实现 1 2 3 4 5 /* *********************************************** 6 Author :111qqz 7 Created Time :2016年03月03日 星期四 14时09分08秒 8 File Name :code/cf/problem/548A.cpp 9 ************************************************ */ 10 11 #include <algorithm> 12 #include <cstdio> 13 #include <iostream> 14 #include <cstring> 15 #include <string> 16 #include <cmath> 17 #include <map> 18 19 using namespace std; 20 const int N=1E3+5; 21 char st[N]; 22 int k,len,ave; 23 bool ok(int l,int r) 24 { 25 for (int i = l ; i<=(l+r)/2;i++) 26 if (st[i]!=st[l+r-i]) 27 return false; 28 return true; 29 } 30 31 int main() 32 { 33 cin>>st; 34 cin>>k; 35 len = strlen(st); 36 ave= len/k; 37 if (k*ave!=len){cout<<"NO"<<endl;return 0;} 38 for (int i = 1 ; i<=k;i++ ) 39 { 40 if (!ok((i-1)*ave,i*ave-1)) 41 { 42 cout<<"NO"<<endl; 43 //cout<<i<<endl; 44 return 0; 45 } 46 47 } 48 cout<<"YES"<<endl; 49 50 51 52 return 0; 53 }

B. Tavas and SaDDas time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn’t stop doing that. That’s why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas’ headphones and told him: “If you solve the following problem, I’ll return it to you.” The problem is: You are given a lucky number n. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

wa了两次，原因是在同一个点可能有多个基地。。。 所以用set 是错误的，应该用multiset 然后因为这道题看到了map+set实现离散化的另外一种写法 我的代码： 代码实现 1/************************************************************************* 2 > File Name: code/hdoj/4022.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月01日 星期六 04时37分20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#define y0 abc111qqz 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define tm crazy111qqz 25#define lr dying111qqz 26using namespace std; 27#define REP(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef unsigned long long ULL; 30const int N=2E5+7; 31const int inf = 0x7fffffff; 32 33map<int,int>xmap,ymap; 34multiset<int>x[N]; 35multiset<int>y[N]; 36int main() 37{ 38 int n,m; 39 while (scanf("%d %d",&n,&m)!=EOF) 40 { 41 if (n==0&&m==0) break; 42 for ( int i = 1 ; i <= n ; i++) 43 { 44 x[i].clear(); 45 y[i].clear(); 46 } 47 xmap.clear(); 48 ymap.clear(); 49 int tx,ty; 50 int cntx=0,cnty=0; 51 for ( int i = 1 ; i <= n ; i++ ) 52 { 53 scanf("%d %d",&tx,&ty); 54 if (!xmap[tx]) xmap[tx]=++cntx; 55 if (!ymap[ty]) ymap[ty]=++cnty; 56 x[xmap[tx]].insert(ymap[ty]); 57 y[ymap[ty]].insert(xmap[tx]); 58 } 59 int c,d; 60 set<int>::iterator it; 61 for ( int i = 1; i <= m ; i++ ) 62 { 63 scanf("%d %d",&c,&d); 64 if (c==0) 65 { 66 cout<<x[xmap[d]].size()<<endl; 67 for ( it = x[xmap[d]].begin();it!=x[xmap[d]].end();it++) 68 { 69 y[*it].erase(xmap[d]); 70 } 71 x[xmap[d]].clear(); 72 } 73 else 74 { 75 cout<<y[ymap[d]].size()<<endl; 76 for ( it =y[ymap[d]].begin();it!=y[ymap[d]].end();it++) 77 { 78 x[*it].erase(ymap[d]); 79 80 } 81 y[ymap[d]].clear(); 82 } 83 } 84 printf("\n"); 85 } 86 87 return 0; 88}