h d electric

always possible to rearrange the bricks such, and all stacks has the same height.The input is terminated by a set starting with n = 0. This set should is processed.Outputfor each set, first print the number of the and the set, as shown in the sample output. Then print the line "the minimum number of moves is K.", where k is the minimum number of bricks that has to be moved I n order to make all the stacks the same height.Output a blank line after each set.Sample Input65 2 4 1 7 50Sample OutputS

2 60 1 2Total 10 2929/10=2.92.9 is the GPA of a bornHere you write a program for calculating GPA.Input contains more than one set of data, and the first row of each group of data has a number n, and the next row of n rows represents a score. Each line has two actual number S,p,s represents the course's credit, p indicates the student's score (percentile). If p=-1 The student This course lacks the examination, should not be counted. Output outputs a row for each set of data, representing the stu

1005ac+ Problem Solving Ideas Analysis: 1, in the title (1 2, (A * f (n-1) + B * F (n-2)) MoD 7 = (a%7*f (n-1) +b%7*f (n-2))%7 3, because F (i) and F (i+1) have only 49 combinations, because (f (i), (i+1) There are only 7 options, that is, only one of the 0,1,2,3,4,5,6. Therefore, the cycle Solving: 1, the cycle is first, by the way, the first period of F (n) to find out. 2, using the cycle, the remaining f (n) is directly obtained. Code: #include #define M 52 in

Eat candyTime limit:6000/3000 MS (java/others) Memory limit:65535/32768 K (java/others)Total submission (s): 27178 Accepted Submission (s): 7691Problem Descriptionhoho, finally from Speakless hand to win all the candy, is Gardon eat candy when there is a special hobby, is not like the same candy put together to eat, like to eat first, the next time to eat another, so But Gardon don't know if there is a candy-eating order so he can finish all the candies? Please write a program to help calculate

问题描述 输入两点坐标(X1,Y1),(X2,Y2),计算并输出两点间的距离。输入 输入数据有多组,每组占一行,由4个实数组成,分别表示x1,y1,x2,y2,数据之间用空格隔开。输出 对于每组输入数据,输出一行,结果保留两位小数。样例输入 0 0 0 1 0 1 1 0 样例输出 1.00 1.41 输入两个坐标，计算两点间距离，无限循环；#include#includeint main(){ doublea,b,c,d,m; while(scanf("%lf%lf%lf%lf",a,b,c,d)!=EOF){m=(a-c)*(a-c)+(b-d)*(b-d);printf("%.2f\n",sqrt(m));}return0; } 1 Note: 1 while(scanf("%lf%lf%lf%lf",a,b,c,d)!=EOF)：实现无限循环！ Two point distance Hangzhou

Verifying the Angle Valley conjectureTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 6653 Accepted Submission (s): 3417There are many speculations in problem description number theory that have not yet been resolved, one of which is known as the "Angular Valley conjecture", a problem that was popular in several prestigious universities in cert, the problem is described as: any one greater than the natural number, if it is odd, is multiplied by

Lowest BitTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 9052 Accepted Submission (s): 6656Problem Descriptiongiven an positive integer a (1 For example, given a = +, we can write a in binary form as 11010, so the lowest bit of a is ten, so the output should be 2 .Another example goes like this:given a = to, we can write A in binary form as 1011000, so the lowest bit of The output should is 8. Inputeach line of input contains only an integer

Eddy ' s-ITime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 7117 Accepted Submission (s): 4268Problem Descriptioneddy ' s interest is very extensive, recently he's interested in prime number. Eddy Discover the all number owned can is divided into the multiply of prime number, but he can ' t write program, so EDD Y have to ask intelligent him, he asks you to write a program which can do the number to divided into the multi Ply of prime number fac

(a[1] = = N) return;if (a[deep]==n) {///when child nodes traverse to N, backtrack a[deep-1]++;D FS (deep-1,n,k,i);} To the fourth layer, if there is a solution output and return, no solution continues to the right if (deep = = 4) {if ((a[1]+a[2]+a[3]+a[4]) = = N) {System.out.println ("Case #" +i+ ":" +a[1]+ "" +a[2]+ "+a[3]+" "+a[4"); return;} else{a[deep]++;D FS (deep,n,k,i);}} If the condition is satisfied then the search is not satisfied, then the pruning if (A[deep] Hangzhou

Find your present!Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 3096 Accepted Submission (s): 2051Problem DescriptionIn The New Year party, everybody'll get a "special present". Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of the them would be yo Urs. Each present have a card number on it, and your present ' s card number would be the one that different from all the others. For examp

A CTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 4206 Accepted Submission (s): 2671Problem Descriptionare You excited the title "AC"? If The answer is YES, AC it;You must learn these and combination formulas in the school. If you had forgotten it, see the picture.Now I'll give you N and M, and your task was to calculate the answer.Inputin the first line, there is a integer T indicates the number of the test cases.Then T-cases follows in the

To find the product of odd numbersTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 56573 Accepted Submission (s): 36490Problem Descriptiongive you n integers and ask for the product of all the odd numbers in them.InputThe input data contains multiple test instances, one row per test instance, and the first number of each row is n, indicating that there is a total of n for this group of data, followed by n integers, you can assume that each set o

) {Set_outh (Gpioa,din); } Else{set_outl (Gpioa,din); } Data=data1; Delay_ms (Ten); Set_outh (GPIOA,SCLK); Delay_ms (Ten); Set_outl (GPIOA,SCLK); Delay_ms (Ten); Set_outh (GPIOA,SCLK); Delay_ms (Ten); } set_outh (Gpioa,din); Delay_ms (Ten);} U32 Readad7715data (U8byte) {U8 i; U32 Data=0; for(i=0;ibyte*8; i++) {Set_outh (GPIOA,SCLK); Delay_ms (Ten); Set_outl (GPIOA,SCLK); Delay_ms ( -); Data=data1; if(read_in (gpioa,dout) = =1) {Data=data+1; } set_outh

The Euler functionTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 4559 Accepted Submission (s): 1902Problem DescriptionThe Euler function phi is a important kind of function in number theory, (n) represents the amount of The numbers which is smaller than N and coprime to n, and this function have a lot of beautiful characteristics. Here comes a very easy question:suppose is given a, B, try to calculate (a) + (a+1) +....+ (b)Inputthere is severa

a few transforms now let you ball some Ball State:Two operations:T a B means to add all the balls in the city where a ball is located to the city where B-ball is locatedQ A indicates the city in which the output a is located , the number of balls in the city wherea is located,the number of times a is transferredAnalysis:Can think of is to use and check set, the first two problems are very good solution, one to find the root node another find the total number of elements in this tree, difficult

from S to T, then output-1.Sample Input3 30 1 10 2 31 2 10 23 10 1 11 2Sample Output2-1AuthorlinleSource2008 Zhejiang University Postgraduate second-round warm-up (2)--Full true simulationObservation point: Shortest Path (DIJKSTRA,FLOYD,SPFA)Main topic: Now there are n towns M Road, give the starting point and end point, ask whether can reach, if can, find the shortest road lengthIn fact, this is a typical shortest path problem, I now Master is three methods, DIJKSTRA,SPFA and Floyd, of which F

; Tim_timebasestructure.tim_countermode=tim_countermode_up; Tim_timebaseinit (TIM2,tim_timebasestructure); // //PA6--->50%//?????Tim_ocinitstructure.tim_ocmode=tim_ocmode_pwm1; Tim_ocinitstructure.tim_outputstate=tim_outputstate_enable; Tim_ocinitstructure.tim_pulse=5;//Pulse???? Period??????Tim_ocinitstructure.tim_outputstate=tim_outputstate_enable; Tim_ocinitstructure.tim_ocpolarity=Tim_ocpolarity_high; Tim_oc1init (TIM2,tim_ocinitstructure); // //PA7--->40%//tim_ocinitstructure.tim_pulse=40

#include 　　Hangzhou Electric 1002

[J] = Dis2[temp] +Mpa[temp][j]; - } the } + } A } the intMain () + { - while(~SCANF ("%d%d", m, n), m|N) $ { $ for(i=1; i) - for(j=1; j) - { theMap[i][j]= (i==j?0: INF); -Mpa[i][j]= (i==j?0: INF);Wuyi } the intA, B, D, p; - for(i =1; I ) Wu { -scanf"%d %d%d%d", a, b, d, p); About /*if (Map[a][b] > D)//pit.! $ Map[a][b]=map[b][a]=d; - if (Mpa[a][b] > P) - mpa[a][b]=mpa[b][a]=p; */ - if

Problem number Fd Type Game Distribution Eventually 1001 5343 SAM W 1002 5344 YES √ 1003 5345 Game W 1004 5346 Probability H 1005 5347 YES √ 1006 5348 YES √ 1007 5349 YES √ 1