general electric application

Fibbonacci numberTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 17675 Accepted Submission (s): 8422Problem Descriptionyour objective for this question are to develop a program which would generate a Fibbonacci number. The FIBBONACCI function is defined as such:F (0) = 0F (1) = 1F (n) = f (n-1) + f (n-2)Your program should is able to handle values of N in the range 0 to 50.Inputeach test case consists of one, integer n in a, where 0≤n≤50. The

Problem description to find the number of n least common multiple.Input inputs contain multiple test instances, and each test instance starts with a positive integer n, followed by n positive integers.Output outputs their least common multiple for each set of test data, with one row for each test instance output. You can assume that the final output is a 32-bit integer.#include intMain(){intN,I,A,S[ -]; while(scanf("%d",N)!=Eof) {A=0; for(I=1;IN;I++) {scanf("%d",S[I]);if(AS[I])A=S[I]; } for(I

The problem is to calculate the number of lattice points inside the triangle, you can use the peak theorem, s = n + b/2-1; where S is the area of the lattice polygon, n is the number of lattice points inside the polygon, and B is the lattice point on the boundary, assuming there are two coordinates (x1, y1) (x2, y2) b = gcd (ABS (X1-X2), (Y1-y2)). The code is as follows:/*id:m1500293 lang:c++ Prog:fence9*/#include#include#includeusing namespacestd;intN, M, p;structpoint{intx, y; Point () {}, poi

Test instructions: Draw a triangle on the checkered paper to find the number of lattice points contained within the triangleBecause one of the edges is the x-axis, the first thing to think about is to figure out the corresponding mathematical function for two edges, and then enumerate the x-coordinate values for solving. But it doesn't have to be that troublesome.Pique theorem: given vertex coordinates are simple polygons of the whole point (or square lattice), the pique theorem illustrates the

Eddy ' s pictureTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 8107 Accepted Submission (s): 4106Problem Descriptioneddy begins to like painting pictures recently, he's sure of himself to become a painter. Every day Eddy draws pictures in his small, and he usually puts off his newest pictures to let his friends appreciate . But the result of it can be imagined, the friends is not a interested in its picture. Eddy feels very puzzled,in order

11 11 2 11 3 3 11 4 6) 4 11 5 10 10 5 1Input data contains multiple test instances, and each test instance input contains only one positive integer n (1Output corresponds to each input, export the Yang Hui triangle of the corresponding number of layers, separated by a space between the integers of each layer, and a blank line behind each Yang Hui triangle.Sample Input2 3Sample Output11 111 11 2 1#include intMain(){intN,I,J,A[ to][ to]; while(scanf("%d",N)!=Eof) { for(I=0;IN;I++) { for(J=0;JI;

Problem description counts the number of Chinese characters in a given text file.The input file first contains an integer n, which indicates the number of test instances, followed by the N-segment text.Output for each piece of text, outputs the number of characters in it, and the output of each test instance takes one row.[Hint:] From the characteristics of the Chinese character machine internal code to consider ~Sample input2wahaha! wahaha! This year the festival does not speak to speak only Pu

program is in the standard output. For each input data set, prints on a, "YES" if it is possible to form a committee and "NO" otherwise . There should not being any leading blanks at the start of the line.An example of program input and output:Sample INPUT2 3 3 3 1 2 3 2 1 2 1 1 3 3 2 1 3 2 1 3 1 1Sample Outputyes NOSourcesoutheastern Europe 2000Recommendwe has carefully selected several similar problems for you:1068 2444 1150 1281 1054 to find the minimum, point coverage (maximum match value)

, boss can make a damage of (b_atk-w_def) to Warrior if (b_atk-w_def) bigger than zero, otherwise no damage. Outputfor each case, if boss's HP first turns to being smaller or equal than zero, please print "Warrior wins". Otherwise, please print "Warrior loses". If Warrior cannot kill the boss forever, please also print "Warrior loses".Sample InputW100 900100 900b100 1000 900100 1000 900Sample Outputwarrior Winswarrior losesWater problem, how much time to find out, pay attention to the case of 0

process it.Outputfor Each test case, the your program should the output one section. The first line of all sections must be ' test Case #k ', where k is the number of the the ' Test case ' (starting with 1). The second one must be ' total explored area:a ', where A is the ' total explored area ' i.e. the area of the the Union of all Rec Tangles in this test case), printed exact to both digits to the right of the decimal point.Output a blank line after each test case.Sample Input210 10 20 2015 1

, the vowels is ' a ', ' e ', ' I ', ' o ', and ' u '; all other letters is consonants.) Note that these rules is not perfect; There is many common/pronounceable words that is not acceptable.Inputthe input consists of one or more potential passwords, one per line, followed by a line containing only the word ' end ' That signals the end of the file. Each password are at least one and at the most twenty letters long and consists only of lowercase letters.Outputfor each password, output whether or

Text ReverseTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 17702 Accepted Submission (s): 6717Problem Descriptionignatius likes to the write words in reverse. Given a single line of text which are written by Ignatius, you should reverse all the words and then output them. Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.Each test case contain

Problem-solving ideas: Because food can be casually eaten, so is a complete backpack, calorie represents consumption, happiness represents value, set of formula can be done.Problem Description for foodies, the happiest thing about the new Year is to eat, not one! 　　But for girls, calories are predators. Senior beauty is familiar with the "fat to fall like a mountain, fat to go like a ladder," the truth, so she hoped you can help develop a recipe, can make her eat happy at the same time, will not

sequence of the first and the end elements.Sample Input6-2 11-4 13-5-210-10 1 2 3 4-5-23 3 7-2165-8 3 2 5 01103-1 -5-23-1 0-20 Sample Output1310 1 410 3 510 100-1 -20 0 0HintHintHuge input, scanf is recommended.The subject uses the idea of dynamic programming, defines 3 arrays, namely DP (state), a (record data), S (record start subscript);The state transition equation is: Dp[i]=max (Dp[i-1]+a[i],a[i]);Core code:for (int i=1;i{if (Dp[i-1]+a[i]>a[i]){Dp[i]=dp[i-1]+a[i];S[I]=S[I-1];}Else{Dp[i]=a[

6del 3sum6add 1add 3add 5add 7add 9sumSample Output345HintC + + maybe run faster than g++ in this problem. The topic has been tangled for a long time, the line tree version of the online problem is not very detailed, it may be strange I can not understand .... Violence vector Step-by-step simulation like can pass, muddle through, after the line tree level deepened later, and then back to write a line tree version hope ~ Learn to debug code with vim for the first time.　Feel OK ~ Just a lot of ma

));Memset(Dp,0 ,sizeof( Dp));S[0]=S[1]=1; for (int I=1;IK;I++) {Cin>>A[I]; } for (int I=1;IK;I + +) { if( Dp[I-1]+A[I]>=A[I]) {Dp[I]=Dp[I-1]+A[I];S[I]=S[I-1]; } Else { Dp[I]=A[I];S[I]=I; }} int Start=1,End=1; int Max=Dp[1]; for (int I=2;IK;I + +) { if( MaxDp[I]) {Max=Dp[I];Start=S[I];End=I; } }cout"Case"Count":"Endl;coutMax" "Start" "EndEndl; if (T!=0)coutEndl; } return 0;} Copyright NOTICE: This article for Bo Master original article, without Bo Master permi

Solution idea: Because the state transfer equation for the complete backpack F[v]=max (F[v],f[v-c[i]]+w[i]) has documented all of the backpack composition scheme, but usually ask for the maximum value, now requires the total number of programsThat is F[v]=sum (F[v],f[v-c[i]+w[i]]),Problem description in a country only 1 points, 2 points, 3 cent coins, the exchange of money n into coins there are many kinds of laws.Please compile the procedure to calculate the total number of different methods.In

The title means: Ask N! How many zeros are there in the tail.Just started talking nonsense, in the last paragraph of the problem description to understand the topic to do.Ask N! The tail of the number of 0, first calculate n! , one more number, it is impossible, and N Max reaches 100000000. And we need to analyze from a mathematical point of view, how is 0 produced?By writing out the factorial of the previous few numbers, you know that to generate 0, you have to have 5 and an even number to mult

Problem-Solving ideas: Direct set formula can be done by a complete backpack.Problem description The undead Lich King, the Death Knight got an n-dollar bill (remember, only a note), in order to prevent his frequent death in the battle, he decided to buy himself some props, so he came to the Goblin store before.Death Knight: "I want to buy props!"Goblin businessman: "We have three kinds of props here, 150 pieces of blood bottle, a magic medicine 200 pieces of one, invincible potion 350 pieces of

]==true ) { while( K!=0) {Sum+=K%Ten;K/=Ten; } if( Prime[Sum]!=true) {Beauty[I]=false; } }}} voidDp()///Dynamic planning idea, dp[i] that is, the number of primes 1 to I {Memset(Dp,0 ,sizeof( Dp)); for (int I=1;I1000000;I + +) { if( Beauty[I]==true) {Dp[I]=Dp[I-1]+1; } Else Dp[I]=Dp[I-1]; }}int main() { int A,B,T,_count;Findprime();Beautyprime();Dp(); while (scanf("%d",T)!=Eof) {_count=0; while (T--) {_count++;scanf("%d%d",A,B);Printf("Case #%d:%d\n",_co