BZOJ2154: Crash的数字表格
题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2154
题意&&题解:http://www.cnblogs.com/jianglangcaijin/archive/2013/11/27/3446169.html
我只是来发泄的!!!怎么每次打数论题都会被爆int,爆long long这种问题纠缠好久!!!!!!!!!!
我还以为这题是双倍经验呢
代码:
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 }