bzoj2820: YY的GCD

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2820

题意：中文题。。。

分析：同bzoj2301，论文题。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=10000010;
const int MAX=151;
const int MOD=1000000007;
const int MOD1=100000007;
const int MOD2=100000009;
const int INF=2100000000;
const double EPS=0.00000001;
typedef long long ll;
typedef unsigned long long uI64;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int a
,bo
,mu
,su
;
ll sum
;
void deal() {
int i,j,k,w,n=10000000;
memset(bo,0,sizeof(bo));
mu[1]=1;sum[0]=k=0;
for (i=2;i<=n;i++) {
if (!bo[i]) { a[++k]=i;mu[i]=-1; }
for (j=1;j<=k;j++) {
if (a[j]*i>n) break ;
bo[a[j]*i]=1;
if (i%a[j]==0) { mu[a[j]*i]=0;break ; }
mu[a[j]*i]=-mu[i];
}
}
memset(su,0,sizeof(su));
for (i=1;i<=k;i++) {
w=a[i];
while (w<=n) { su[w]+=mu[w/a[i]];w+=a[i]; }
}
for (i=1;i<=n;i++) sum[i]=sum[i-1]+su[i];
}
ll get(int n,int m) {
ll ret=0;
int i,last;
if (n>m) { n^=m;m^=n;n^=m; }
for (i=1;i<=n;i=last+1) {
last=min(n/(n/i),m/(m/i));
ret+=(ll)(n/i)*(m/i)*(sum[last]-sum[i-1]);
}
return ret;
}
int main()
{
int n,m,t;
scanf("%d", &t);
deal();
while (t--) {
scanf("%d%d", &n, &m);
printf("%lld\n", get(n,m));
}
return 0;
}

/*
10
10 10
100 100
*/