general electric blender

This question, simple minimal spanning tree problem.Just enter the time is more troublesome, the beginning of N, is the number of villages, the following N-1 information, the beginning of the capital letter S and K, is the S village has a k road connection, followed by the K village and the weight value.It's easy to handle the data you've entered.The following is the code for the AC:#include Hangzhou Electric Acm1301--jungle roads~~ minimum spanning t

The wrong answer started, it took one hours to find out what was wrong and I was drunk. (Please note after the loop to clear 0, on more than once when the)#include #include int main (){int n,m,i,j,sum=0,num=0;while (scanf ("%d%d", n,m)!=eof){if (n{for (i=n;i{if (i%2==0){Sum+=i*i;}Else{Num+=i*i*i;}}printf ("%d%d\n", sum,num);}Else{for (i=m;i{if (i%2==0){Sum+=i*i;}Else{Num+=i*i*i;}}printf ("%d%d\n", sum,num);}}}Positive solution#include #include int main (){int n,m,i,j,sum=0,num=0;while (scanf ("%

, -, to, -, to};3 intJudgeintYear )4 {5 if(Year%4==0 Year% -!=0) || Year% -==0)6 return 1 ; 7 Else8 return 0 ;9 }Ten intMain () One { A intI, J, N, y, M, D; - -scanf"%d", n); the while(n--) - { -scanf"%d-%d-%d", y, m, d); - if(M = =2 judge (y) d = = in!judge (y+ -))//No 18-year-old birthday; + { -printf"-1\n") ; + Continue ; A } at intsum =0 ; - for(i = y; i -; i++) - { -Sum + =365 ; - if

; thein4=1; the Break; About Case 0x43://Braking thein1=0; theIn2=0; thein3=0; +in4=0; - Break; the Case 0x07: Break;//eq for return key, return back to first functionBayi Case 0x15: Break; the Case 0x09: Break; the Case 0x16: Break; - Case 0x19: Break; - Case 0x0d: Break; the Case 0x0c: Break; the Case 0x18: Break; the Case 0x5e:

Len is constant, that is, the longest increment of the subsequence is unchanged, then replaced with the smaller element, is to makeFurther elements are more "opportunity" to add to the DP "" at the end of the 1, in other words, the potential for longer sequences increases.Also pay attention to the output format details. Road and roads. Also note that the end of the binary search must be low above high and per "low". Y is greater than the number you are looking forAC:#include Hangzhou

); - intLen =strlen (str); - intA=0, b=0, c=0, d=0, e=0 ; the for(i=0; i) - { - if(Str[i] = ='a') -a++ ; + if(Str[i] = ='e') -b++ ; + if(Str[i] = ='I') AC++ ; at if(Str[i] = ='o') -d++ ; - if(Str[i] = ='u') -e++ ; - } -printf"a:%d\n", a); inprintf"e:%d\n", b); -printf"i:%d\n", c); toprintf"o:%d\n", d); +printf"u:%d\n", e); - if(n!=0) theprintf"\

This question, looked for a long time, did not find what the law, Baidu after, only know is the number of the smallest permutation is the first of the enumeration.What a long knowledge!! ~Knowing this rule, you can quickly write, algorithm this header file in another enumeration of the function next_permutation, the smallest permutation of the first, is called Next_permutation I-1 times after the arrangement . The same applies to next_permutation, which means that there are duplicate elements in

The main problem is to find the law, the first time to find out! ~~The topic is to ask for a number of digital root, this so-called digital root is a number of the sum of the numbers, if the number of double digits, repeat the digital root, until the number is a number.This problem is to seek n^n 's digital root.The rules are as follows:The result of N-n multiplication is assumed to be the multiplication of the digital root of s,s, which is equal to the number of n digital root.The following is

The problem is the simple application of the minimal spanning tree. At first, did not think of using the smallest spanning tree to do, think of is greedy, when known to use the smallest spanning tree to do, but also made a very serious mistake, is the time complexity of the estimate wrong, leading to the beginning dare not write, thinking of other methods. Think of it as a lesson.The following is the code of the AC, with detailed comments, with a check set to judge the ring, the time complexity

program should print exactly one blank line.Sample Input3 5 the - 63923 99999Sample Output 3 5 Good Choice 20 63923 99999 Good ChoiceConsider a large number; Pay attention to the format;1#include 2 #defineMAX 100010;3 4 intgcdintAintb//greatest common divisor;5 {6 if(!B)returnA;7 Else8 returnGCD (b,a%b);9 }Ten One intMain () A { - intM,n; - while(~SCANF ("%d%d",m,N)) the { - intlen=gcd (m,n); - if(len==1) -printf"%10d%10d G

This problem, simple BFS can be done. The approximate meaning of the topic is the shortest time from the r of the map to reach a.In the beginning, with the normal queue to do, the result of memory is super, the reason is N and M maximum 200; a normal queue wastes a lot of memory, so you should use a priority queue instead.The following is the code for AC:#include Hangzhou Electric acm1242--rescue~~bfs+ Priority queue

U Calculate ETime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 35178 Accepted Submission (s): 15843Problem Descriptiona Simple Mathematical formula-E iswhere n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of N.Outputoutput The approximations of e generated by the above formula for the values of nFrom 0 to 9. The beginning of your output should appear similar to that shown be

just because of one detail wrong wa one afternoon. now talk about this problem, generally we bfs when the depth and adjacency table related, this because the focus is the direction, we can think of the turn 0 bends is a layer, 1 bends is the second floor .... In accordance with this rule, BFs arrives at eachthe number of turns in the junction is the fewest number of times, I was wrong in the beginning of the program logic is not a search in the end. Instead of encountering a node that has been

Lowest Common Multiple PlusTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 39183 Accepted Submission (s): 16144Problem 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 t

Number of Split planes = number of intersections + vertex number +1Make f (n-1) The number of planes divided by the first n-1 polyline, when the nth polyline is added.Because each edge intersects with the two edges of the front n-1 polyline, the number of intersections is increased by 2*2* (n-1) and the vertex is increased by 1, soF (n) =f (n-1) +4 (n-1) +1F (n-1) =f (n-2) +4 (n-2) +1....F (2) =f (1) +4*1+1F (1) =2F (N) -2=4 ((n-1) + (n-2) +...+1) + (n-1) =4* ((1+n-1) * (n-1)/2) +n-1=2 (n*n-n) +

low energy particle quantity. Each output occupies one row.Sample Input5 2 -1Sample Output571 209 One 4 long int against GNU C + +, using __int64 against VC6SOURCE2006/1/15 ACM Program Design Final Exam//careful;1#include 2__int64 biao[ *],sieve[ *];3 intMain ()4 {5 intI,j,n;6biao[1]=3; sieve[1]=1;7 for(i=2;i the; i++)8 {9biao[i]=biao[i-1]*3+sieve[i-1]*2;Tensieve[i]=biao[i-1]+sieve[i-1]; One } A while(~SCANF ("%d", n) n!=-1) - { - if(n==0) theprintf"1, 0\n");

First, the ADC conceptTo convert analog signals into digital signals. Second, the Code framework#include"stm32f10x.h"voiddelay (u32 kk) { while(kk--);}intMain () {U16 AD=0, adc0=0; Gpio_inittypedef gpio_initstructure; Adc_inittypedef adc_initstructure; Rcc_apb2periphclockcmd (Rcc_apb2periph_gpioa|RCC_APB2PERIPH_ADC1, ENABLE); Gpio_initstructure.gpio_pin=gpio_pin_0; Gpio_initstructure.gpio_mode=Gpio_mode_ain; Gpio_init (Gpioa,gpio_initstructure); Adc_initstructure.adc_mode=adc_mode_independent;

Peach in mindTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 41117 Accepted Submission (s): 30915Problem DescriptionLike the journey of the students must know that Wu empty eat peach story, you must all think this monkey too noisy, in fact you are some do not know: Goku is a mathematical problem in the study!What's the problem? The question he studied was how many peaches there were!However, in the end, he still failed to solve the problem, heh

Problem 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 the first line of input has an integer t, followed by the T Group of data, each group of data accounted for 2 rows, the first row is an integer

guaranteed.The so state transfer equation is: Dp[i]=max (Dp[i-1]+a[i], a[i]), Time complexity O (n^2), successful acceptCode#include #include #include using namespace Std;int array1[100002];int dp[100002];int main () {int k;Cin >> K;int f=0;while (k--){int A;Cin >> A;int start=1;int end=1;memset (array1,0,sizeof (array1));Memset (Dp,0,sizeof (DP));int sum=-9999;int i;for (i=1;i{CIN >> Array1[i];Dp[i]=max (Dp[i-1]+array1[i],array1[i]);if (dp[i]>sum) {//used to update sumSum=dp[i];End=i;}}int sum