Acdream 1148 GCD sum Mobius Inversion ansx, ansy

GCD sumtime limit: 8000/4000 ms (Java/others) memory limit: 128000/64000 KB (Java/others) submitstatisticnext problemproblem description gives n, m
Run the following program:
Long long ans = 0, ansx = 0, ansy = 0;
For (INT I = 1; I <= N; I ++)
For (Int J = 1; j <= m; j ++)
If (gcd (I, j) = 1) ans ++, ansx + = I, ansy + = J;
Cout <ans <"" <ansx <"" <ansy <Endl; Input

Multiple groups of data, with N and M (1 <= n, m <= 100000) in each row ).

 

Output is as described in the question. Each row outputs three numbers, which are separated by spaces: ANS, ansx, and ansy.

5 51 3

Sample output

19 55 553 3 6

The total hint data is less than 50000. It is easier for the first digit to use Mobius for inversion. For ansx, it is similar to ansy. Now we will discuss ansx practices. let's set F (d) To represent the sum of X when 1 <= x <= N, 1 <= Y <= m, gcd (x, y) = D. when we set F (d) to represent 1 <= x <= N, 1 <= Y <= m, and gcd (x, y) to a multiple of D, so F (1) = F (1) + F (2) + f (3 )...... F (min (n, m) F (2) = F (2) + f (4) + f (6 ).... F (min (n/2, M/2); f (I) = f (I) + f (I * 2) + f (I * 3 ).... F (min (N/I, m/I); therefore, we can find that due to d = 1, we can filter Mu [I] * I based on the inversion, calculate the sum of the First n items. Use parts to complete the process. Code:

 1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 typedef long long LL; 7  8 const int maxn = 1e5+3; 9 bool s[maxn];10 int prime[maxn],len = 0;11 int mu[maxn];12 LL hxl [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         }25         for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++)26         {27             s[i*prime[j]] = false;28             if(i%prime[j]!=0)29                 mu[i*prime[j]] = -mu[i];30             else31             {32                 mu[i*prime[j]] = 0;33                 break;34             }35         }36     }37     for(int i=1;i<maxn;i++)38         sum1[i] = sum1[i-1]+mu[i];39     hxl[1] = mu[1];40     for(int i=2;i<maxn;i++){41         hxl[i] = i*mu[i]+hxl[i-1];42     }43 }44 int main()45 {46     init();47     int n,m;48     while(scanf("%d%d",&n,&m)>0)49     {50         LL sum = 0;51         LL ansi = 0,ansj = 0;52         int a = n;53         int b = m;54         if(a>b) swap(a,b);55         for(int i=1,la = 0;i<=a;i++,i = la+1)56         {57             la = min(a/(a/i),b/(b/i));58             sum = sum + ((LL)(a/i))*(b/i)*(sum1[la]-sum1[i-1]);59             ansi = ansi +(hxl[la]-hxl[i-1])*(((LL)(n/i+1)*(n/i))/2)*(m/i);60             ansj = ansj +(hxl[la]-hxl[i-1])*(((LL)(m/i+1)*(m/i))/2)*(n/i);61         }62         printf("%lld %lld %lld\n",sum,ansi,ansj);63     }64     return 0;65 }

 

Acdream 1148 GCD sum Mobius Inversion ansx, ansy