BZOJ2154: Crash的数字表格

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 10000000+5
 26 
 27 #define eps 1e-10
 28 
 29 #define ll long long
 30 
 31 #define pa pair<int,int>
 32 
 33 #define for0(i,n) for(int i=0;i<=(n);i++)
 34 
 35 #define for1(i,n) for(int i=1;i<=(n);i++)
 36 
 37 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 38 
 39 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 40 
 41 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 42 
 43 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
 44 #define mod 20101009
 45 
 46 using namespace std;
 47 
 48 inline int read()
 49 
 50 {
 51 
 52     int x=0,f=1;char ch=getchar();
 53 
 54     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 55 
 56     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 57 
 58     return x*f;
 59 
 60 }
 61 int p[maxn],tot,mu[maxn];
 62 bool v[maxn];
 63 void get(int m)
 64 {
 65     mu[1]=1;
 66     for2(i,2,m)
 67     {
 68         if(!v[i])p[++tot]=i,mu[i]=-1;
 69         for1(j,tot)
 70         {
 71             int k=i*p[j];
 72             if(k>m)break;
 73             v[k]=1;
 74             if(i%p[j])mu[k]=-mu[i];
 75             else {mu[k]=0;break;}
 76         }
 77     }
 78     for1(i,m)mu[i]=((ll)mu[i]*i%mod*i%mod+mu[i-1])%mod;
 79 }
 80 inline int sum(int n,int m)
 81 {
 82     return ((ll)n*(n+1)/2%mod)*((ll)m*(m+1)/2%mod)%mod;
 83 }
 84 inline int f(int n,int m)
 85 {
 86     int ret=0;
 87     if(n>m)swap(n,m);
 88     for(int i=1,j;i<=n;i=j+1)
 89     {
 90         j=min(n/(n/i),m/(m/i));
 91         ret=(ret+(ll)(mu[j]-mu[i-1])*sum(n/i,m/i)%mod)%mod;
 92     }
 93     return ret;
 94 }
 95 
 96 int main()
 97 
 98 {
 99 
100     freopen("input.txt","r",stdin);
101 
102     freopen("output.txt","w",stdout);
103         int n=read(),m=read(),ans=0;get(m);
104         if(n>m)swap(n,m);
105         for(int i=1,j;i<=n;i=j+1)
106         {
107             j=min(n/(n/i),m/(m/i));
108             ans=(ans+(ll)(i+j)*(j-i+1)/2%mod*f(n/i,m/i)%mod)%mod;
109         }
110         printf("%d\n",(ans+mod)%mod);
111 
112     return 0;
113 
114 }