题目大意：略
果然是一道神duliu题= =
出题人的题解还是讲得很明白的
1.关于$sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}varphi (i,j)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}frac{varphi (i)varphi (j)gcd(i,j)}{varphi (gcd(i,j))}$的证明，lgl神犇提供了一种方法
假设现在$gcd(i,j)$中有一个质因子$p$，幂次是$k$，那么它对$varphi (gcd(i,j))$的贡献就是$p^{k-1}(p-1)$
设$i$中$p$的幂次是$k_{1}$，$j$是$k_{2}$。则他们对$varphi (i),varphi (j)$的贡献分别是$p^{k_{1}-1}(p-1)$和$p^{k_{2}-1}(p-1)$，相乘得$p^{k_{1}+k_{2}-2}(p-1)^{2}$
而$p$对$varphi (ij)$的贡献是$p^{k_{1}+k_{2}-1}(p-1)$，所以$frac{p^{k}}{p^{k-1}(p-1)}=frac{p}{p-1}$
$frac{p}{p-1}*p^{k_{1}+k_{2}-2}(p-1)^{2}=p^{k_{1}+k_{2}-1}(p-1)$
而质因子间互不影响，因此得证
2.对于最后化简出来的式子$sumlimits_{Q=1}^{n} left ( sumlimits_{d|Q}frac{d}{varphi (d)}mu(frac{Q}{d}) ight ) left ( sumlimits_{i=1}^{left lfloor frac{n}{Q} ight floor} varphi (iQ) ight ) left ( sumlimits_{i=1}^{left lfloor frac{m}{Q} ight floor} varphi (jQ) ight )$ 的处理
整除分块不能同时处理两个定义域不同的函数相乘！
所以要把第一个式子$left ( sumlimits_{d|Q}frac{d}{varphi (d)}mu(frac{Q}{d}) ight )$ 按照常规方法进行整除分块，第二个式子$left ( sumlimits_{i=1}^{left lfloor frac{n}{Q} ight floor} varphi (iQ) ight ) $和$left ( sumlimits_{i=1}^{left lfloor frac{m}{Q} ight floor} varphi (jQ) ight )$用官方题解的方法处理
蒟蒻用$vector$写的常数巨大
1 #include <vector> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define N1 100010 6 #define ll long long 7 #define dd double 8 using namespace std; 9 10 const int B=31; 11 const int maxn=100000; 12 const int jr=998244353; 13 14 void exgcd(ll a,ll b,ll &x,ll &y) 15 { 16 if(!b){ x=1; y=0; return; } 17 exgcd(b,a%b,x,y); ll t=x; x=y; y=t-a/b*y; 18 } 19 int mu[N1],phi[N1],use[N1],pr[N1],cnt; 20 int f[N1],G[N1][B+1][B+1]; 21 vector<int>g[N1]; 22 void init() 23 { 24 int i,j,Q,n; mu[1]=phi[1]=1; 25 for(i=2;i<=maxn;i++) 26 { 27 if(!use[i]){ pr[++cnt]=i; mu[i]=-1; phi[i]=i-1; } 28 for(j=1;j<=cnt&&i*pr[j]<=maxn;j++) 29 { 30 use[i*pr[j]]=1; 31 if(i%pr[j]){ mu[i*pr[j]]=-mu[i]; phi[i*pr[j]]=phi[i]*phi[pr[j]]; } 32 else{ phi[i*pr[j]]=phi[i]*pr[j]; break; } 33 } 34 } 35 int ans=0;ll inv,y; 36 for(Q=1;Q<=maxn;Q++) 37 { 38 ans=0; 39 for(i=1;i*Q<=maxn;i++) 40 { 41 ans=(phi[i*Q]+ans)%jr; 42 g[Q].push_back(ans); 43 } 44 } 45 for(i=1;i<=maxn;i++) 46 { 47 exgcd(phi[i],jr,inv,y); inv=(inv%jr+jr)%jr; 48 for(j=1;j*i<=maxn;j++) 49 f[i*j]=(1ll*i*inv%jr*mu[j]%jr+f[i*j]+jr)%jr; 50 //sf[i]=(sf[i-1]+f[i])%jr; 51 } 52 for(Q=1;Q<=maxn;Q++) 53 { 54 for(i=1;i<=B;i++) for(j=1;j<=B;j++) 55 G[Q][i][j]=(1ll*g[Q][i-1]*g[Q][j-1]%jr*f[Q]+G[Q-1][i][j])%jr; 56 } 57 } 58 int n,m,T; 59 60 int main() 61 { 62 scanf("%d",&T); 63 init(); 64 int i,j,la;ll ans=0; 65 while(T--) 66 { 67 scanf("%d%d",&n,&m); if(n>m) swap(n,m); 68 for(i=1,ans=0;i<=n&&m/i>B;i++) 69 ans=(ans+1ll*g[i][n/i-1]*g[i][m/i-1]%jr*f[i]%jr)%jr; 70 for(;i<=n;i=la+1) 71 { 72 la=min(n/(n/i),m/(m/i)); 73 ans=(ans+G[la][n/i][m/i]-G[i-1][n/i][m/i]+jr)%jr; 74 } 75 printf("%lld ",ans); 76 } 77 return 0; 78 }