2818: Gcd
Time Limit: 10 Sec Memory Limit: 256 MBSubmit: 1566 Solved: 691
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Description
给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的
数对(x,y)有多少对.
Input
一个整数N
Output
如题
Sample Input
4
Sample Output
4
HINT
hint
对于样例(2,2),(2,4),(3,3),(4,2)
1<=N<=10^7
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 7 typedef long long LL; 8 const int maxn = 1e7+1; 9 bool s[maxn]; 10 int prime[maxn],len = 0; 11 int mu[maxn]; 12 int g[maxn]; 13 int sum1[maxn]; 14 void init() 15 { 16 memset(s,true,sizeof(s)); 17 mu[1] = 1; 18 for(int i=2; i<maxn; i++) 19 { 20 if(s[i] == true) 21 { 22 prime[++len] = i; 23 mu[i] = -1; 24 g[i] = 1; 25 } 26 for(int j=1; j<=len && (long long)prime[j]*i<maxn; j++) 27 { 28 s[i*prime[j]] = false; 29 if(i%prime[j]!=0) 30 { 31 mu[i*prime[j]] = -mu[i]; 32 g[i*prime[j]] = mu[i] - g[i]; 33 } 34 else 35 { 36 mu[i*prime[j]] = 0; 37 g[i*prime[j]] = mu[i]; 38 break; 39 } 40 } 41 } 42 for(int i=1; i<maxn; i++) 43 sum1[i] = sum1[i-1]+g[i]; 44 } 45 46 int main() 47 { 48 int a; 49 init(); 50 while(scanf("%d",&a)>0) 51 { 52 LL sum = 0; 53 for(int i=1,la = 0 ; i<=a; i = la+1) 54 { 55 la = a/(a/i); 56 sum = sum + (long long)(sum1[la] - sum1[i-1])*(a/i)*(a/i); 57 } 58 printf("%lld ",sum); 59 } 60 return 0; 61 }
spoj
4491. Primes in GCD TableProblem code: PGCD |
Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has created a GCD (greatest common divisor) table! So he now has a table (of height a and width b), indexed from (1,1) to (a,b), and with the value of field (i,j) equal to gcd(i,j). He wants to know how many times he has used prime numbers when writing the table.
Input
First, t ≤ 10, the number of test cases. Each test case consists of two integers, 1 ≤ a,b < 107.
Output
For each test case write one number - the number of prime numbers Johnny wrote in that test case.
Example
Input:
2
10 10
100 100
Output:
30
2791
一样的题，只不过 GCD(x,y) = 素数 . 1<=x<=a ; 1<=y<=b;
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 7 typedef long long LL; 8 const int maxn = 1e7+1; 9 bool s[maxn]; 10 int prime[maxn],len = 0; 11 int mu[maxn]; 12 int g[maxn]; 13 int sum1[maxn]; 14 void init() 15 { 16 memset(s,true,sizeof(s)); 17 mu[1] = 1; 18 for(int i=2;i<maxn;i++) 19 { 20 if(s[i] == true) 21 { 22 prime[++len] = i; 23 mu[i] = -1; 24 g[i] = 1; 25 } 26 for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++) 27 { 28 s[i*prime[j]] = false; 29 if(i%prime[j]!=0) 30 { 31 mu[i*prime[j]] = -mu[i]; 32 g[i*prime[j]] = mu[i] - g[i]; 33 } 34 else 35 { 36 mu[i*prime[j]] = 0; 37 g[i*prime[j]] = mu[i]; 38 break; 39 } 40 } 41 } 42 for(int i=1;i<maxn;i++) 43 sum1[i] = sum1[i-1]+g[i]; 44 } 45 46 int main() 47 { 48 int T,a,b; 49 init(); 50 scanf("%d",&T); 51 while(T--) 52 { 53 scanf("%d%d",&a,&b); 54 if(a>b) swap(a,b); 55 LL sum = 0; 56 for(int i=1,la = 0 ;i<=a;i = la+1) 57 { 58 la = min(a/(a/i),b/(b/i)); 59 sum = sum + (long long)(sum1[la] - sum1[i-1])*(a/i)*(b/i); 60 } 61 printf("%lld ",sum); 62 } 63 return 0; 64 }