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, but, think about it, 25 is definitely a multiple of 5, so, the above has been counted once, so only need to add 80 is good.80/5 = 16 (equivalent to 2015/125), is the 1~2015 between 16 125 multiples, plus 16, for the same reason.16/5 = 3 (equivalent to 2015/625), is the 1~2015 between 3 625 multiples, plus 3, for the same reason.3/5 = 0 (equivalent to 2015/3125); not a multiple of 3125.So 2015! There are 403 + 80 + 16 + 3 consecutive 0 in the back.The following is the code for the AC:#include

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

{ for(int J=2*I;J1000000;J+=I) {Prime[J]=false;Beauty[J]=false; }}//prime number filter core code area}void Beautyprime () { int Sum,K; for (int I=2;I1000000;I++) {Sum=0;K=I; if (Prime[I]==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; }

number 1 or 0, or 1 if you are the winner, or 0 for the other.Sample Input2 18 44 7Sample Output010SourcenoiRecommendll | We have carefully selected several similar problems for you:1404 1536 1517 1524 1729 Witzov game (Wythoff game): There are two piles of various items, two people take turns From a heap or at the same time from the two piles to take the same number of items, the provisions of at least one at a time, more than open, the last to win the light.In this case it is quite complicate

Problem number Hdu Type Game Distribution Eventually 1001 5327 YES √ 1002 5328 YES √ 1003 5329 H 1004 5330 W 1005 5331 H 1006 5332 W 1007 5333 H 1008 5334

The meaning of the title is: the Rabbit casually hide in a hole, the wolf from the hole No. 0 began to find, every time to find the next is a number of M, all the holes formed ring.Input m and n, Hole is number No. 0 to n-1.The greatest common divisor of M and N is 1, and the wolf can be found in every hole. Otherwise, some holes are ignored, and those holes are safe.The following is the code for the AC:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master per

, the value is 2.For each golden stick, the value is 3.Pudge wants to know the total value of the hook after performing the operations.Consider the original hook is a made up of cupreous sticks.Inputthe input consists of several test cases. The first line of the input is the number of the cases. There is no more than cases.For each case, the first line contains a integer N, 1Next Q lines, each line contains three integers X, Y, 1Outputfor, print a number in a line representing the total value of

Problem description for a given string, statistics the number of occurrences of a numeric character.Input data has multiple lines, the first line is an integer n, indicating the number of test instances, followed by n rows, each containing a string consisting of letters and numbers.Output for each test instance, outputs the number of values in the string, with each output occupying one row.Sample INPUT2ASDFASDF123123ASDFASDFASDF111111111ASDFASDFASDFSample Output69Code:1#include 2#include malloc.

Nuclear reactorTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 12219 Accepted Submission (s): 5547Problem description There are two types of events occurring in a nuclear reactor:When the high-energy particle strikes the nucleus, the particle is absorbed, releasing 3 high-energy particles and a low-energy particle.When the low-energy particle strikes the nucleus, the particle is absorbed, releasing 2 high-energy particles and a low-energy par

one row.Sample Input 5 2-1Sample Output 571, 20911, 4Tip You can use long long int against GNU C + + and use __int64 against VC6AuthorUnknownSource 2006/1/15ACM Program Design Final ExamRecommendLcyAnalysis: From the No. 0 second to 1 seconds, 1 high-energy particles emit 3 high-energy particles and a low-energy particle, from the 1th second to 2nd seconds 3 high-energy particles and a low-energy particle generation (3*3+2*1) a high-energy particle, while 3 high-energy particles produce 3 low-e

city of 8600 and the city to goOutput If 86 fails to reach his destination, outputs "What a pity!",The other outputs the safety factor of the safest road between these two cities and retains three decimal places.Sample Input31 0.5 0.50.5 1 0.40.5 0.4 131 22 31 3Sample Output0.5000.4000.500authorailyanlu the same two times, each time has a new feeling: with AC code: #include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Hangzhou

A = = B?Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 64239 Accepted Submission (s): 10060Problem descriptiongive You both numbers A and B, if A is equal to B, you should print "YES", or print "NO".Inputeach test Case contains-numbers A and B.Outputfor each case, if A was equal to B, you should print "YES", or print "NO".Sample Input1 22 23) 34 3Sample OutputNoyesyesno/* The first time to do is always to understand the topic too simple. Aft

= Find (root*2, left, right); $b = Find (root*2+1, left, right); - returnMax (A, b); - } the intMain () - {Wuyi while(~SCANF ("%d%d", n, m)) the { - for(inti =1; I ) Wuscanf"%d", num[i]); -Build (1,1, N); About Charstr[Ten];intA, B; $ for(inti =1; I ) - { -CIN >> Str >> a >>b; - if(str[0] =='Q') Aprintf"%d\n", Find (1, A, b)); + Else the { -Num[a] =b; $Update (1, A, b); the } the } the

subtree, which will be used in later lookups.The picture is stolen, there is no time to do it yourself.2, find: Find the number of k on the interval [a, b]. If the number of children entering the left child is greater than or equal to K, go to the left subtree to find, if less than, then go to the right subtree lookup, find the number of k-s large, s for the interval [a, b] into the number of left subtree. Update interval [A, b].The code can be used as a template.#include Copyright NOTICE: Thi

This problem, similar to the decryption of ciphertext. The meaning of the topic is very clear.Give you the number of P Q e l,l for the second row.Calculate the N and FN first, and then calculate the D.n = p * q;fn = (p-1) * (q-1).Then according to the D * e% fn = 1, calculate d.The key is to seek C. The number num to you is equal to c ^ d% n. c The corresponding ASCLL code is the decrypted character.The following is the code for the AC:#include Hangzhou Elec