hdu 5663 Hillan and the girl 莫比乌斯反演
Hillan and the girl
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Problem Description
“WTF!
While everyone has his girl(gay) friend, I only have my keyboard!”
Tired of watching others' affair, Hillan burst into scream, which made
him decide not to hold it back.
“All right, I am giving you a question. If you answer correctly, I will be your girl friend.” After listening to Hillan, Girl replied, “What is the value of ∑ni=1∑mj=1f, where f if gcd(i,j) is a square number and f if gcd(i,j) is not a square number(gcd(i,j) means the greatest common divisor of x and y)?”
But Hillan didn't have enough Intelligence Quotient to give the right answer. So he turn to you for help.
“All right, I am giving you a question. If you answer correctly, I will be your girl friend.” After listening to Hillan, Girl replied, “What is the value of ∑ni=1∑mj=1f, where f if gcd(i,j) is a square number and f if gcd(i,j) is not a square number(gcd(i,j) means the greatest common divisor of x and y)?”
But Hillan didn't have enough Intelligence Quotient to give the right answer. So he turn to you for help.
Input
The first line contains an integer T——The number of the test cases.
For each test case, the only line contains two integers n,m(1≤n,m≤10,000,000) with a white space separated.
For each test case, the only line contains two integers n,m(1≤n,m≤10,000,000) with a white space separated.
Output
For each test case, the only line contains a integer that is the answer.
Sample Input
2
1 2333333
10 10
Sample Output
0
33
Hint
In the first test case, obviously $f\left(i,j\right)$ always equals to 0, because $i$ always equals to 1 and $\gcd\left(i,j\right)$ is always a square number(always equals to 1).
Source
min(n,m) min(n/k,m/k)
思路:首先推到∑ ∑ mu(d) * [n/k/d] * [m/k/d]; k为完全平方数;
k=1 d=1
令T=k*d;
可得:
min(n,m)
∑ [n/T] * [m/T] ∑ mu(T/k) ;
T k|T
令gg数组等于 ∑ mu(T/k) 相当于原来的mu函数;
k|T
和原来一样分块即可;
#include<bits/stdc++.h> using namespace std; #define ll __int64 #define esp 0.00000000001 #define pi 4*atan(1) const int N=1e7+10,M=1e7+10,inf=1e9+10,mod=1e9+7; int mu[N], p[N], np[N], cnt, sum[N]; ll gg[N]; void init() { mu[1]=1; for(int i=2; i<N; ++i) { if(!np[i]) p[++cnt]=i, mu[i]=-1; for(int j=1; j<=cnt && i*p[j]<N; ++j) { int t=i*p[j]; np[t]=1; if(i%p[j]==0) { mu[t]=0; break; } mu[t]=-mu[i]; } } for(int i=1;i*i<N;i++) { for(int t=i*i;t<N;t+=(i*i)) sum[t]+=mu[t/i/i]; } for(int i=1;i<N;i++) gg[i]=gg[i-1]+sum[i]; } ll getans(ll b,ll d) { if(b>d)swap(b,d); ll ans=0; for(ll L=1,R=0;L<=b;L=R+1) { R=min(b/(b/L),d/(d/L)); ans+=(b/L)*(d/L)*(gg[R]-gg[L-1]); } return ans; } int main() { int T; init(); scanf("%d",&T); while(T--) { ll b,d; scanf("%I64d%I64d",&b,&d); printf("%I64d\n",(b*d)-getans(b,d)); } return 0; }