best number theory books

Hdoj 1286 find new friends [Euler's function of number theory], hdoj1286 Find new friends Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission (s): 7912 Accepted Submission (s): 4157 Problem Description the New Year is approaching. The pig Association is preparing to have a party. We already know that there are N members and the

is the same as the result of the previous ac .... However, I am forced to use judge as a judge because I cannot use another AC code pair. If it is illegal, the output should be set as INF, then, the maximum value I set is different from the INF value set by the AC code. I thought I was wrong and debug for several days. Today, I found that the slot ,,,, in fact, I may always be right. Well, of course, I will talk about it later when uvalive is ready. The idea is to reverse the prime

The main topic: given n bottles, choose K, you can guide the oil casually, asked to choose K bottles can be exported to the minimum value of the maximum number of oilFirst of all, K bottles can be exported to the minimum oil must be K bottle capacity of the largest common factor so the problem is converted to the n number of the choice K to maximize the maximum common factorFind out all the factor of n

Problem 1"The main topic"Give theMultiple sets of data, given in order to find out. SolvingProof: In addition to think that all are even, so the number of coprime is paired.It's up to you.So for each pair and for, there is a common pair.TheProblem 2"The main topic"Write on the first circle, write on the second circle, and then each time on the basis of the previous circle, write their and each of the two numbers, defined as the

because of the traversal of the residual series produced by the mutual quality. Same Theorem 6: If A, B, C, and D are four integers, and a then B (mod m), C then D (mod m), then AC then BD (mod m ); Understanding: A can be divided into mk1 + H forms, C can also be divided into Mk2 + H forms, and then (mk1 + H) (mk3 + l) Merge (Mk2 + H) (mk4 + l) (mod m) Proof of ferma's theorem: If P is a prime number, there must be a

Max FactorTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d %i64 U DescriptionTo improve the organization of he farm, Farmer John labels each of its n (1 (Recall that a prime number is just a number, the has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by

, "/stack:102400000,102400000")intScan () {intres =0, ch; while( ! (ch = getchar ()) >='0' CH '9' ) ) { if(ch = = EOF)return 1 - ; } Res= CH-'0' ; while(ch = getchar ()) >='0' CH '9') Res= Res *Ten+ (CH-'0' ) ; returnRes;} VectorV;vectorans;ll prime[maxn],cnt;BOOLVIS[MAXN];voidPrime () {CNT=0; memset (Vis,0,sizeof(VIS)); V.push_back (Make_pair (1LL,1LL)); for(LL i=2; i) { if(!Vis[i]) {prime[cnt++]=i; for(DoubleJ= (Double) i*i;ji) V.push_back (Make_pair (LL) j,i); }

Test instructions: A number in addition to all factors other than itself, the resulting number if less than itself, output deficient, otherwise output abundant, if equal, output perfectThe core of the problem: to ask for all the factors other than itselfint sum=1, I; for (i=2; i) { if(n%i==0) sum+ =i; }A nu