PoPoQQQ讲义第4题
题解：http://www.cnblogs.com/jianglangcaijin/archive/2013/11/27/3446169.html
感觉两次sqrt(n)的枚举是亮点……
RE：汗- -b 10^7是8位数，开数组少打了一个0……
1 /************************************************************** 2 Problem: 2154 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:8780 ms 7 Memory:167292 kb 8 ****************************************************************/ 9 10 //BZOJ 2154 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<iostream> 15 #include<algorithm> 16 #define rep(i,n) for(int i=0;i<n;++i) 17 #define F(i,j,n) for(int i=j;i<=n;++i) 18 #define D(i,j,n) for(int i=j;i>=n;--i) 19 using namespace std; 20 21 int getint(){ 22 int v=0,sign=1; char ch=getchar(); 23 while(ch<'0'||ch>'9') {if (ch=='-') sign=-1; ch=getchar();} 24 while(ch>='0'&&ch<='9') {v=v*10+ch-'0'; ch=getchar();} 25 return v*=sign; 26 } 27 /*******************tamplate********************/ 28 const int N=10000086,P=20101009; 29 typedef long long LL; 30 LL prime[N],mu[N]; 31 bool check[N]; 32 LL n,m; 33 void getmu(int n){ 34 int tot=0; 35 mu[1]=1; 36 for(int i=2;i<n;i++){ 37 if (!check[i]){ 38 prime[tot++]=i; 39 mu[i]=-1; 40 } 41 rep(j,tot){ 42 if (i*prime[j]>n) break; 43 check[i*prime[j]]=1; 44 if (i%prime[j]) mu[i*prime[j]]=-mu[i]; 45 else{ 46 mu[i*prime[j]]=0; 47 break; 48 } 49 } 50 } 51 F(i,1,n) mu[i]=(mu[i-1]+mu[i]*i%P*i%P)%P; 52 } 53 inline LL Sum(LL n,LL m){ 54 n=n*(n+1)/2%P; 55 m=m*(m+1)/2%P; 56 return n*m%P; 57 } 58 inline LL f(LL n,LL m){ 59 LL ans=0,i,last; 60 for(i=1;i<=n;i=last+1){ 61 last=min(n/(n/i),m/(m/i)); 62 ans=(ans+(mu[last]-mu[i-1])%P*Sum(n/i,m/i)%P)%P; 63 } 64 return ans; 65 } 66 int main(){ 67 // freopen("input.txt","r",stdin); 68 n=getint(); m=getint(); 69 if(n>m) swap(n,m); 70 getmu(m); 71 LL ans=0,i,last; 72 for(i=1;i<=n;i=last+1){ 73 last=min(n/(n/i),m/(m/i)); 74 ans=(ans+(i+last)*(last-i+1)/2%P*f(n/i,m/i)%P)%P; 75 } 76 if (ans<0) ans+=P; 77 printf("%lld\n",ans); 78 return 0; 79 }