assassin creed pc

The Assassin's Creed Mutiny will land on the PC platform on March 24, and the minimum configuration requirements have been given to the authorities. Configuration requirements: Operating system: Windows 7 SP1 or Windows 8/8.1 (64-bit)CPU: Intel dual-core Q6600@2.4ghz or AMD Fast Dragon II X4 620@2.6ghzRam:2gbGraphics card: NVIDIA GeForce GTS450 or AMD Radeon HD5670 (1G video memory) or Intel HD4600Direc

() {//p->p+top; intnum =0, rest =m; for(inti =0; I 0]; ++i) {if(Rest >= p[0][i].need) {num++; Rest-= p[0][i].need; } Else Break; } returnPII (num, M-rest);} PII worktwo () {intrest = m, num =0; int from[2]; from[0] = from[1] =0; if(top[1] ==0|| Rest 1][0].need)returnPii0,0); Rest-= p[1][0].need; from[1]++; BOOLFlag; for (;;) { if(N-( from[0]+ from[1]) can) Break; Flag=false; if( from[1] 1] p[0][ from[0]].need > p[1][ from[1]].need) {if(Rest >= p[1][ from[1]

HDU 4415 Assassin's Creed, hdu4415 It's hard to think, greedy. Others' ideas: Click to open the link Problem DescriptionEzio Auditore is a great master as an assassin. now he has prowled in the enemies 'base successfully. he finds that the only weapon he can use is his cuff sword and the sword has durability m. there are n enemies he wants to kill and killing each enemy needs Ai durability. every time Ezio

First in My Computer-documents, find the Assassin's Creed document file (Note: path D:/documents/assassin ' s Creed Rogue) Click on the file opening assassin There is a Notepad file, open, in the contents of the file to find Betop the beginning of the file, as shown and change it to the numbers I circle

durability of the smallest kill, the answer to add the number of swordsIf you can't kill a sword, don't kill it:Residual durability is to kill the minimum durability, pay attention to the first to kill the possible inside of the sword, it will skip;　　1#include 2#include 3 using namespacestd;4 #defineN 1000105 structenemy{intb;} T[n];6 BOOLCMP (enemy X,enemy y) {returnx.ay.a;}7 intMain () {8 intt,n,m,cas=1;9scanf"%d",T);Ten while(t--){ One inti,cost=0, num=0, k=-1; Ascanf"%d%d",

It's hard to think, greedy. Others' ideas: Click to open the link Problem descriptionezio Auditore is a great master as an assassin. now he has prowled In the enemies 'base successfully. he finds that the only weapon he can use is his cuff sword and the sword has durability M. there are n enemies he wants to kill and killing each enemy needs AI durability. every time Ezio kills an enemy he can use the enemy's sword to kill any other bi enemies without

A2.We also found that B1 cannot be replaced because it does not have a father.So the positive solution:#include #include#includeusing namespacestd;#defineN 100050#definell Long Longintt,n,ac,bc;ll ax[n],bx[n],by[n],m;BOOLCMP (ll x,ll y) {returnxy;}intCS;intMain () {scanf ("%d",T); while(t--) {CS++; printf ("Case %d:", CS); scanf ("%d%i64d",n,m); ll Sy=1; AC=bc=0; for(intI=1; i) {BC++; scanf ("%i64d%i64d",bx[bc],BY[BC]); if(!BY[BC]) {ax[++ac]=BX[BC]; BC--; Continue; } sy+=BY[BC]; } sort (Ax+1,

What should I do if I receive a 0xc0000005 error from the Assassin's Creed in Russia?Problem:My name is win7, and all other components supporting games are installed. I don't know why this problem occurs.Solution:Right-click the game execution file and select the third "compatibility troubleshooting" to enable the system to automatically detect available compatibility and save the settings

Source: Light OJ 1406 Assassin ' s Creed Test instructions: There is a way to send the fewest people through all the cities and everyone can't walk where others walk. Idea: the least of the people can walk the complete graph is obviously the smallest path coverage problem there may be a ring so to shrink point but look at the sample and find that a strong connected component may be split n Max 15 so state c

Source: Light OJ 1429 Assassin ' s Creed (II) Test instructions: At least a few people walk the full picture can repeat the graph Idea: If it is a DAG diagram and each point cannot be repeated then the bare minimum path overlay is now not a DAG may have loops and each point may be repeated for a graph after a ring can be indented to a point where the figure is a DAG graph another point can be repeated walk