标签：mil rac find 测试数据 repr except origin euc example
新年快到了，“猪头帮协会”准备搞一个聚会，已经知道现有会员N人，把会员从1到N编号，其中会长的号码是N号，凡是和会长是老朋友的，那么该会员的号码肯定和N有大于1的公约数，否则都是新朋友，现在会长想知道究竟有几个新朋友？请你编程序帮会长计算出来。Input第一行是测试数据的组数CN（Case number，1<CN<10000），接着有CN行正整数N（1<n<32768），表示会员人数。Output对于每一个N，输出一行新朋友的人数，这样共有CN行输出。
Sample Input
2 25608 24027
Sample Output
7680 16016
#include"stdio.h" typedef long long ll; int main() { ll i,j,phi[100000],n,m; for(i=2;i<100000;i++) phi[i]=0;phi[1]=1; for(i=2;i<100000;i++) { if(phi[i]==0) { for(j=i;j<100000;j+=i) { if(phi[j]==0) phi[j]=j; phi[j]=phi[j]*(i-1)/i; } } } scanf("%lld",&n);i=0; while(i++<n) { scanf("%lld",&m); printf("%lld\n",phi[m]); } }
Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!
InputInput contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.
OutputFor each integers N you should output the number of integers M in one line, and with one line of output for each line in input.
Sample Input
2 4 0
Sample Output
0 1
#include"stdio.h" typedef long long ll; int main() { ll i,j,n,ret,m; while(scanf("%lld",&n)!=EOF) { if(n==0)break; if(n==1) { printf("0\n");continue; } ret=n;m=n-1; for(i=2;i*i<=n;i++) { if(n%i==0) { ret=ret-ret/i; n=n/i; while(n%i==0) n=n/i; } } if(n>1) ret=ret-ret/n; printf("%lld\n",m-ret); }
Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.InputFor each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.OutputFor each test case, you should print the sum module 1000000007 in a line.Sample Input
3 4 0
Sample Output
0 2
#include"stdio.h" typedef long long ll; int main() { ll i,j,n,ret,m; while(scanf("%lld",&n)!=EOF) { if(n==0)break; ret=n;m=n; for(i=2;i*i<=n;i++) { if(n%i==0) { ret-=ret/i; n=n/i; while(n%i==0) n=n/i; } } if(n>1) ret=ret-ret/n; printf("%lld\n",((m-1)*m/2-ret*m/2)%1000000007); } }
The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+....+ (b)InputThere are several test cases. Each line has two integers a, b (2<a<b<3000000).OutputOutput the result of (a)+ (a+1)+....+ (b)Sample Input
3 100
Sample Output
3042
#include"stdio.h" typedef long long ll; int main() { ll i,j,n1,n2,phi[3000001]={0}; for(i=2;i<3000001;i++) phi[i]=0; for(i=2;i<3000001;i++) { if(phi[i]==0) { for(j=i;j<3000001;j+=i) { if(phi[j]==0) phi[j]=j; phi[j]=phi[j]*(i-1)/i; } } phi[i]=phi[i-1]+phi[i]; } while(scanf("%lld%lld",&n1,&n2)!=EOF) { printf("%lld\n",phi[n2]-phi[n1-1]); } }
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.InputThe first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.OutputFor each test case,output the answer on a single line.Sample Input
3 1 1 10 2 10000 72
Sample Output
1 6 260
标签：mil rac find 测试数据 repr except origin euc example
原文地址：http://www.cnblogs.com/lch316/p/6768151.html
