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Topic Portal1 /*2 Water questions: three strings to determine if each has a corresponding vowel letter, yes/no3 Afternoon Slow speed: (4 */5#include 6#include 7#include string>8#include 9#include Ten#include One using namespacestd; A - Const intMAXN = 1e2 +Ten; - Const intINF =0x3f3f3f3f; the strings[3]; - - intMainvoid)//codeforces Beta Round #70 (Div. 2) A.

Topic Portal1 /*2 Shortest way: Not only scan the edges, but also scan the points; There are two cases, one is just at the midpoint, that is, from the u,v, then the last/23 there is a different from the u,v, both of the distance is L4 The template was wrong and it was a long time: (5 */6#include 7#include 8#include 9#include Ten#include One#include A using namespacestd; - - Const intMAXN = 1e5 +Ten; the Const intINF =0x3f3f3f3f; - intD[MAXN]; - int

Codeforces Round #275 (Div. 2 ), Link: http://codeforces.com/contest/483 A. CounterexampleTime limit per test1 secondmemory limit per test256 megabytes Your friend has recently learned about coprime numbers. A pair of numbers {A, Bytes,B} Is called coprime if the maximum number that divides bothAAndBIs equal to one. Your friend often comes up with different statements. He has recently supposed that if th

Codeforces Round #279 (Div. 2) Problem Solving report A. B .C.D. E A-Team Olympus IAD Greedy questions .. You can start from the first one. The Code is as follows: # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include using n

A problem-solving ideas: that is, given the shape of 1,1+2,1+2+3,1+2+3+4,----, 1+2+3+4+---The +n of the series, give you a number to judge it in the first few items of the series. The first n entries and formulas for this sequence are S (n) =n* (n+1) * (n+2)/6, proof s (n) =n* (n+1) * (n+

() {intN, I, POS, flag; while(scanf("%d", n)!=eof) { for(i=1; iscanf("%d", a[i]); }if(A[n]) {puts("NO");Continue; }if(n==1){puts("yes\n0\n");Continue; }if(a[n-1]){puts("yes\n"); for(i=1; iprintf("%d", A[i]);if(i!=n)printf(" ,"); }Continue; }if(n==2){puts("NO");Continue; } flag=0; for(i=n-2; i>=1; i--) {if(!a[i]) {flag=1; Pos=i; Break; } }if(!flag) {puts("NO");Continue; }puts("YES"); for(i=1; iprintf

Codeforces Round # Pi (Div. 2) C. Geometric Progression dpC. Geometric Progression time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarp loves geometric progressions very much. Since he was only three years old, he loves only the progressions of length three. He also has a favorite integerKAnd a sequenceA, ConsistingNIntegers. He wants to know

, A +1+N, CMP1);//for (int i = 1; I //cout // }St.push (n); intAns =0; for(inti = n-1; I >=1; --i) {if(a[i].a[2] = = A[i +1].a[2]) {St.push (i); intres = Ask (A[i].id +1); Ans+ = res > a[i].a[3]; } Else { while(!St.empty ()) { intres =St.top (); St.pop (); UpDate (a[res].id, a[res].a[3]); } intres = Ask (A[i].id +1); Ans+ = res > a[i].a[3]; St.push (i

Codeforces Round # Pi (Div. 2) (STL Session ), Codeforces Round # Pi (Div. 2) A-Lineland Mail Questions, fight for speed. /** @ Author Novicer * language: C ++/C */# include B-Berland National Library "-X" that has not appeared at the beginning of preprocessing. Add "+ x" at the beginning" Then, use set to simulat

codeforces Round #489 (Div. 2)A. Nastya and an Array#include #define REP (i,a,b) for (int i=a;i#define FREP (I,A,B) for (int i=a;i>=b;--i)#define MEM (w) memset (w,0,sizeof (w) )#define PB Push_backtypedef Long LongllConst intN = -;Const intINF =0x3f3f3f3f;inline intRead () {CharC=getchar ();intx=0, f=1; while(c' 0 '|| C>' 9 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 'c' 9 ') {x=x*Ten+c-' 0 '; C=getcha

Topic Portal1 /*2 Test Instructions: gives a series of records of a reader's travel, + denotes a reader to enter,-indicates that a reader has left, and may have had readers in the library before3 Construction: Now records the current number of libraries, SZ records the smallest capacity, in the array tagged readers, in the case of a discussion4 */5 /************************************************6 * Author:running_time7 * Created time:2015-8-6 0:23:3

Codeforces Round #Pi (Div. 2)A-lineland MailWater problem, hand speed./** @author novicer* language:c++/c*/#include B-berland National LibraryPre-processing "-X" not appearing at the beginning, supplementing "+ X" at the beginningThen use set to simulate the process of entering and leaving, output the maximum set capacity./** @author novicer* language:c++/c*/#include C-geometric progressionEnumerate each el

Topic Portal1 /*2 Test instructions: gives a non-circular graph, the degree of each point and the xor of the adjacent points (a^b^c^ ...)3 graph theory/bit operations: In fact, the problem is very simple. Similar topological sort, first in the degree of 1 first-in pairs, each one less than one degree4 the key is to update the XOR, the essence: a ^ b = c, a ^ c = b, b ^ c = A;5 */6#include 7#include 8#include 9#include Ten#include One using namespaces

Topic Portal1 /*2 Test Instructions: Ask if you can use the mass of w^0,w^1,..., w^100 weights each weigh m, weights left or on the right3 Brute Force/conversion: Suppose it can be called, denoted by W, each of which is 0,1,w-1. W-1 means that weights and items are put together, and the simulation determines whether each bit is OK4 Detailed Explanation:http://blog.csdn.net/u011265346/article/details/465563615 Summary: The game did not go into the syst

Topic Portal1 /*2 Test instructions: N digital turntable, starting with each turntable pointing to a number (0~n-1, counter-clockwise sort), and then every turn, odd +1, even 1, ask how many times to make the first I digital turntable point to i-13 Construction: The first to ask the 1th point to 0 how many steps, according to the number of times after the ability to meet the requirements4 the topic reads well tired: (5 */6#include 7#include 8#include

Baby Ming and Weight liftingTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)Total submission (s): 681 Accepted Submission (s): 280Problem Descriptionbaby Ming is fond of weight lifting. He has a barbell pole (the weight of which can being ignored) and a different kinds of barbell disks (the weight of which are respectivelya and b), the amount of each one being infinite.Baby Ming prepare to use this, kinds of barbell disks to make up a new one weighted C(The Barbe

Topic Portal1 /*2 construction: From large to small structure, each time the last is not 9 into 9,p-p MOD 10^k-1, until less than the minimum value. 3 In addition, up to len-1 cycles4 */5#include 6#include 7#include 8#include 9 using namespacestd;Ten OnetypedefLong Longll; A Const intMAXN = 1e3 +Ten; - Const intINF =0x3f3f3f3f; - the intGet_len (ll x) { - intRET =0; - while(x) { -X/=Ten; ret++; + } - returnret; + } A at intGet_nine

Codeforces Round #455 (Div. 2) C. Python Indentation Test instructions: In Python, given n for loops or statements, ' F ' must have a statement. Combine the python indentation into a legitimate program and ask how many possible scenarios. Tags:dp DP[I][J] Indicates the number of possible scenarios at which the I-statement indent is J, and the transfer:

Codeforces Round #277.5 (Div. 2) partial question A. SwapSorttime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output In this problem your goal is to sort an array consistingNIntegers in at mostNSwaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are saved med consecutively, one after another. Note

; - vec[i].clear (); - } - for(i=2; i) - { inscanf"%d",n); - Vec[n].push_back (i); to } + Long LongT=DFS1 (1); -dp[1]=1.0; theDfs1);p rintf ("%.1f", dp[1]); * for(i=2; i) $ {Panax Notoginsengprintf"%.1f", Dp[i]); - } theprintf"\ n"); + } A return 0; the } + Long L