SPOJ4491. Primes in GCD Table（gcd(a,b)=d素数，（1<=a<=n,1<=b<=m））加强版

SPOJ4491.PrimesinGCDTable Problemcode:PGCD

Johnnyhascreatedatablewhichencodestheresultsofsomeoperation--afunctionoftwoarguments.Butinsteadofaboringmultiplicationtableofthesortyoulearnbyheartatprep-school,hehascreatedaGCD(greatestcommondivisor)table!Sohenowhasatable(ofheightaandwidthb),indexedfrom(1,1)to(a,b),andwiththevalueoffield(i,j)equaltogcd(i,j).Hewantstoknowhowmanytimeshehasusedprimenumberswhenwritingthetable.

Input

First,t≤10,thenumberoftestcases.Eachtestcaseconsistsoftwointegers,1≤a,b<107.

Output

Foreachtestcasewriteonenumber-thenumberofprimenumbersJohnnywroteinthattestcase.

Example

Input:

2
1010
100100

Output:

30
2791

一样的题，仅仅只是GCD(x,y)=素数.1<=x<=a;1<=y<=b;链接：http://www.spoj.com/problems/PGCD/转载请注明出处：寻找&星空の孩子具体解释：http://download.csdn.net/detail/u010579068/9034969#include<stdio.h>#include<string.h>#include<algorithm>usingnamespacestd;constintmaxn=1e7+5;typedeflonglongLL;LLpri[maxn],pnum;LLmu[maxn];LLg[maxn];LLsum[maxn];boolvis[maxn];voidmobius(intN){LLi,j;pnum=0;memset(vis,false,sizeof(vis));vis[1]=true;mu[1]=1;for(i=2;i<=N;i++){if(!vis[i])//pri{pri[pnum++]=i;mu[i]=-1;g[i]=1;}for(j=0;j<pnum&&i*pri[j]<=N;j++){vis[i*pri[j]]=true;if(i%pri[j]){mu[i*pri[j]]=-mu[i];g[i*pri[j]]=mu[i]-g[i];}else{mu[i*pri[j]]=0;g[i*pri[j]]=mu[i];break;//think...}}}sum[0]=0;for(i=1;i<=N;i++){sum[i]=sum[i-1]+g[i];}}intmain(){mobius(10000000);intT;scanf("%d",&T);while(T--){LLn,m;scanf("%lld%lld",&n,&m);if(n>m)swap(n,m);LLt,last,ans=0;for(t=1;t<=n;t=last+1){last=min(n/(n/t),m/(m/t));ans+=(n/t)*(m/t)*(sum[last]-sum[t-1]);}printf("%lld\n",ans);}return0;}