fiio x3

size is not reduced. The size is 35x35, the number of channels increases, and the number of 4 branch channels and 64 + 64 + 96 + 32 = 256, the final output tensor size is 35x35x256. 1st Inception Module group: 2nd Inception modules, named Mixed_5c. Step 1: padding model SAME. Four branches, and the fourth branch is connected to 64 output channels with 1x1 convolution. The size of the output tensor is 35x35x288. 1st Inception Module group: 3rd Inception modules, named Mixed_5d. The size of the o

correspondence relationship. The number of bar codes that meet the conditions is the number of k tuples. That is, the number of integer solutions of the equation Σ (I = 1 .. k) xi = n, 1 ≤ xi ≤ m. The easiest way to think of this is to search for algorithm 1: Since the value range of xi has been determined, we can leave all xi values empty and then check how many groups of solutions meet the requirements of Σ (I = 1 .. k) xi = n. The program is easy to write, but the complexity is very high. Fo

. Output* Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows. Sample Input 43 12 52 64 3 Sample Output 57 Source USACO 2004 u s Open A group of cows have different degrees of hearing loss after the festival. When I heard others' speeches, the volume of others must be higher than v [I], when two cattle I and j communicate, the minimum communication sound is max {v [I], v [j]} * the distance between them. Now there are nheaded cows, please ask

# Include # Include # Include Using namespace std;Double segment (double a, double B );Double area (double a, double B, double c );Int main (){Int j, I = 0;Double h [10000];Double x1, y1, x2, y2, x3, y3;Cin> x1> y1> x2> y2> x3> y3;While (x1 + x2 + x3 + y1 + y2 + y3 )! = 0){Double a = segment (x1-x2), (y1-y2 ));Double B = segment (x2-

variable. Similarly, opening different cmd windows has its own character_set_client variable. Lab 3 07/16/2010 1. Create a table with the default Character Set utf8 (navicat, code page 65001 on the utf8 page) and insert the utf8 Encoded chinese character'' 2. Switch to mysql console (code page 936) 3. Set names gbk; then the created table is displayed. Is it true? ---- Yes! Of course, only character_set_results into gbk can be displayed normally. Lab 4 1. Mysql console (code page 936) creat

independent application with its own character_set_client variable. Similarly, opening different cmd Windows has its own character_set_client variable. Lab 3 07/16/2010 1. Create a table with the default character set utf8 (navicat, code page 65001 on the utf8 page) and insert the utf8 encoded Chinese character'' 2. Switch to mysql console (Code page 936) 3. Set names gbk; then the created table is displayed. Is it true? ---- Yes! Of course, only character_set_results into gbk can be displa

Draw a circular multi-angle Pattern Based on html5, and draw a multi-angle Pattern Based on html5 Let's take a look at the simplest: The Code is as follows:Copy the content to the clipboard using JavaScript Code Var canvas = document. getElementById ('My), ctx = canvas. getContext ('2d '); SetInterval (function (){ Ctx. clearRect (0, 0, 400,400 ); Ctx. save (); Ctx. translate (200,200 ); Var ci = 90, pi = Math. PI/ci, x1 = 100, y1 = 0, x2 = 0, y2 = 0,

); return 0 ;}View Code#include #include#includeusing namespace std; int Main () {int n;cin>>n;cout10 )%; return 0 ;}View Code#include #includeusingnamespace std; int Main () { int v,h,r; CIN>>h>>R; V=h*r*r*3.14; cout20000/v+1; return 0 ;}View Code#include #include#includeusing namespace std; int Main () { double xa,ya,xb,yb; CIN>>xa>>xb>>ya>>YB; printf ("%.3lf", sqrt ((Xa-ya) * (Xa-ya) + (XB-YB) * (xb-YB)); return 0 ;}View Code#include #include#includeus

values between 5.0 and 7.5: Here's how to produce a value between 10 5.0 and 7.5: > x2 > 2 > x2 > 2[1] 6.339188 5.311788 7.099009 5.746380 6.720383 7.433535 7.159988 [1] 6.339188 5.311788 7.099009 5.746380 6.720383 7.4335 35 7.159988[8] 5.047628 7.011670 7.030854 [8] 5.047628 7.011670 7.030854 Generate a random integer between 1 and 10 generates a 1 to 10This looks likes the same exercise as the last one, though now we are only want whole numbers, not fractional values. It looks

Title: Give the coordinates of three points, the circumference of the circumscribed circle of the triangle of three points, assuming the known triangle side length a,b,c and triangular area s, then circumscribed circle diameter d = (a*b*c)/(2*s)#include #include#includeusing namespacestd;Const DoublePi=3.141592653589793;intMain () {Doublex1,y1,x2,y2,x3,y3,a,b,c,st,cir,r,p; while(cin>>x1>>y1>>x2>>y2>>x3>>y3

ifABS (x) ThenExit0); //accuracy within 0 output * ifX>0 ThenExit1)ElseExit (-1); $ End;Panax NotoginsengFunction between (Y1,y2,y3:real): boolean; //determine if a point is between two points - Begin the ifThreeout (y1-y3) *threeout (y1-y2) 0 ThenExit (True)Elseexit (FALSE); + End; AFunction Check (X1,y1,x2,y2,x3,y3,x4,y4:real): boolean; //determine if they intersect. the varS1,s2,s3,s4:longint; + begin -S1:=threeout (Cros

returnf*x; - } - voidBuild1 (intx,intLintR) { -X=++size; sz[x]=ch[x][0]=ch[x][1]=0;if(L==R)return; + intMid= (l+r) >>1; Build1 (ch[x][0],L,MID); Build1 (ch[x][1],mid+1, R); - } + voidInsertintx,intLintRintval) { Asize++; sz[size]=sz[x]+1; ch[size][0]=ch[x][0]; ch[size][1]=ch[x][1]; X=size;if(L==R)return; at intMid= (l+r) >>1;if(val0],l,mid,val);if(val>mid) Insert (ch[x][1],mid+1, r,val); - } - voidDfsintXintFA) { -RT[X]=RT[FA]; Insert (Rt[x],1, tot,v[x]); - for(intI=G[X];I;I=ES

Poj_2954 The core of this question is the pick theorem: The given vertex coordinates are simple polygon of the entire point. The Area A and the number of internal points I and the number of edge points B satisfy the following requirements: A = I + B/2-1. Therefore, we can first obtain the area of the triangle and the number of cells on the edge, and then obtain the number of cells in the triangle. You can obtain the number of cells on the edge by finding the maximum number of comments of | D

Topic Link: Click to open the linkThe main topic: there is a wine barrel capacity for VC, there are VW's wine, and now use VB water to brush wine barrels, every time the inner wall of the barrel will leave a VR liquid, up to brush k times, ask how to brush wine barrels, can let the wine barrel inside the least.If VB+VW In other cases, it is guaranteed that the water will be poured out. Therefore, the remaining liquid in the final bucket is VR, as long as the amount of wine in the VR is guarantee

Test instructions, given a string, whether there is a contiguous substring, satisfies that the substring has more than 3 characters, is in ascending order and the number of each character is the same, such as a. Ab... BC...C where a b c is equal in number and is incremented. Because, when more than 3 times, always 3 characters are satisfied, so only consider is the case of 3, Num[i] that the first position, the back up to Num[i] characters are the same. If, there is such a

#includestring.h>#includeusing namespacestd; structNode {intx,y,z,t; }; Const intd[4][3]= {{0,1,0},{0,0,1},{0,-1,0},{0,0,-1}}; Charss[ One][101][101]; intpp[ One][101][101];inth,w,l; ints,v[101],p[101],f[10001]; BOOLzg[ One][101][101]; Charsss[102];node r[100005];intGgg,gg;BOOLCheckintZintXinty) {if(x>-1y>-1z>-1zL)return 1; return 0; } voidBFS (intXxintYyintZzintTT) { inti; intx3,y3,z3; if(ss[zz][xx][yy]=='P') {P[pp[zz][xx][yy]]=tt*3; } if(ss[zz][xx][yy]=='U'zg[zz+1][xx][yy]==0) {Zg

"POJ 2954" TriangleA problem with a very strong pose involves three points.1. Triangle algorithm known three points S = (AB X BC)/2(PS: Helen formula 2. It is known that two points for two points is the maximum common multiple of the two-point transverse ordinate difference ( -1-– does not contain vertices)(GCD (ABS (X2-X1), ABS (Y2-Y1)) (-1))3. The area of the lattice point polygon s = d/2+t-1 (the d-> polygon boundary of the lattice points t-> polygon inner lattice points to find the T-move te

value. At this time, X2 (1) =[3,1.5]; F (X2 (1)) =-7.5. Verify that the terminating iteration condition is met | | X2 (1)-x0 (1) | | = 2.06>err.calculate function descent in all directions:1=f (X0 (1))-F (X1 (1)) =4; 2=f (X1 (1))-F (X2 (1)) =0.5; m=min{4,0.5}=4; map points:X (1) = 2x2 (1)-x0 (1) =2*[3,1.5] '-[1,1]=[5,2]; Conditional Validation: f3=f (X (1)) =-7, F3(F1-2F2+F3) (f1-f2-m) ^2 = 1.25 So, get a new search direction: S (1) =x2 (1)-x0 (1) =[3,1.5]-[1,1]=[2,1.5] ';do a one-dimen

[u]!=Top[v]) - { $ if(d[top[u]]D[top[v]) swap (U,V); theu=Fa[top[u]]; the } the if(d[u]D[v]) swap (U,V); the returnv; - } in intDMinintX1,intX2,intX3)//find the deepest point in depth the { the if(D[x1]>=d[x2]d[x1]>=d[x3])returnX1; About if(D[x2]>=d[x1]d[x2]>=d[x3])returnx2; the returnX3; the } the intDisintUintV//Ask for distance + { - intans=0; the while(top[u]

Follow the previous tutorial to assemble a 3D printer for 800 yuan K800 is a low-cost 3D Printer Based on kosselmini. It reduces the cost by changing the design, but reduces the performance and cost-effectiveness. The main changes are:Change the base bracket to-General BracketChange planetary step to-"proportional gear step"Change the straight-line slide to-> slideFish-eye Joint Parallel arm-"high-intensity parallel arm Motor part:43mm long 42 step motor X4, 42 step for k800 (kosselmini) 3D pri