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canteen meal card has a very strange design, that is, before purchasing to determine the balance. If the remaining amount on the card is greater than or equal to 5 yuan before the purchase of an item, the purchase must be successful (even if the balance is negative on the card after purchase), it cannot be purchased (even if the amount is sufficient). So we all want to try to keep the balance on the card to the minimum. One day, the canteen has n vegetables for sale, each vegetable can be purch

Number SequenceTime limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 11925 Accepted Submission (s): 5440Problem Descriptiongiven-sequences of numbers:a[1], a[2], ..., a[n], and b[1], b[2], ..., b[m] (1 Inputthe first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is the numbers n and M (1 Outputfor Each test case, you should the output one line which only contain K described above. If

from the inside of the structure; to if(a[i].step>d) +ans++; -Dfs (K,a[i].step);//The structure has been zeroed out. the } * $ }Panax Notoginseng - intMain () the { + intT, I, J, X, y; A while(~SCANF ("%d", t)) the { + while(t--) - { $Memset (Head,-1,sizeof(head)); $Memset (A,0,sizeof(a)); -scanf"%d%d", n, d); -Len =0; the for(i=1; i) - {Wuyiscanf"%d%d", x, y); the Add (x, y); - } WuAns =0; -Dfs (0,0); Aboutprint

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

program prints a single real number to both decimal places:the minimum total length of ink lines that can Conne CT all the points.Sample Input31.0 1.02.0 2.02.0 4.0Sample Output3.41Another template problem, the minimum spanning tree Kruskal algorithm, but the premium Manaus algorithm is not very useful, has been used Kruskal, but the premium Manaus algorithm can handle big data, which is presumably excellent: with AC code:#include Copyright NOTICE: This article for Bo Master original article,

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

; - } + intMain () { - intT; +scanf"%d", t); A while(t--){ atscanf"%d%d", n, u); -memset (Map,0,sizeof(map)); -memset (DIS,0,sizeof(DIS)); - for(inti =1; I ){ - intA, B; -scanf"%d", a); in for(intj =1; J ){ -scanf"%d", b); toMAP[I][B] =1; + } - } the intCNT =0; * for(inti =1; I ){ $memset (Vis,0,sizeof(Vis));Panax Notoginseng if(Search (i)) -cnt++; the } +printf (cnt==n?)"yes\n":"no\n"); A }

Warrior and "B" for boss. They attack each and turn.The second line contains three integers, w_hp, w_atk and W_def. (1third line contains three integers, b_hp, b_atk and B_def. (1Note:warrior can make a damage of (w_atk-b_def) to Boss if (w_atk-b_def) bigger than zero, otherwise no damage. Also, 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

## Description: # The people who is crazy enough to think they can change the world, is the ones who does! =============================================================================*/# #include Version 2: /*=============================================================================## Author:liangshu-cbam # # QQ : 756029571 # # School: Harbin Polytechnic University # # last modified:2015-08-11 12:50## filename:a.cpp## Description: # The people who is crazy enough to think they can cha

consecutive occurrences of the same letter, and except for ' ee ' or ' oo '.(For the purposes of this problem, 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 t

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

input sample). If all k elements are negative, define their maximum and 0, outputting the entire 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[

).Output for each operation "sum" tab print one line containing exactly one integer denoting the digest sum of the Curre NT set. Print 0 If the set is empty.Sample Input9add 1add 2add 3add 4add 5sumadd 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 simulati

Main () { int T,K,Count=0;Cin>>T; while (T--) {Cin>>K;Count++;Memset(A,0 ,sizeof( A));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;}

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