bzoj2154: Crash的数字表格

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2154

题意：中文题。。

分析：同bzoj2301，莫比乌斯反演论文题。。为什么我的怎么慢。。20s边缘。不知道200ms的是什么鬼。。。

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 long long MOD=20101009;
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[N],bo[N],mu[N],pre[N];
ll sum[N];
void deal() {
    int i,j,k,w,n=10000000;
    memset(bo,0,sizeof(bo));
    mu[1]=1;k=0;sum[0]=pre[0]=0;
    for (i=2;i<=n;i++) {
        if (!bo[i]) { a[++k]=i;mu[i]=-1; }
        for (j=1;j<=k;j++) {
            if ((ll)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];
        }
    }
    for (i=1;i<=n;i++) {
        pre[i]=(pre[i-1]+i)%MOD;sum[i]=(sum[i-1]+(ll)i*i*mu[i])%MOD;
    }
}
ll suan(int n,int m) {
    ll ret1,ret2;
    if (n&1) ret1=(ll)((n+1)/2)*n;
    else ret1=(ll)(n/2)*(n+1);
    if (m&1) ret2=(ll)((m+1)/2)*m;
    else ret2=(ll)(m/2)*(m+1);
    ret1%=MOD;ret2%=MOD;
    return (ret1*ret2)%MOD;
}
ll calc(int n,int m) {
    int i,last;
    ll ret=0;
    for (i=1;i<=n;i=last+1) {
        last=min(n/(n/i),m/(m/i));
        ret=(ret+(sum[last]-sum[i-1])*suan(n/i,m/i))%MOD;
    }
    return ret;
}
ll get(int n,int m) {
    int i,last;
    ll ret=0;
    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=(ret+(pre[last]-pre[i-1])*calc(n/i,m/i))%MOD;
    }
    return ret;
}
int main()
{
    int i,n,m;
    ll ans=0;
    scanf("%d%d", &n, &m);
    deal();
    ans=get(n,m);
    printf("%lld\n", (ans+MOD)%MOD);
    return 0;
}

/*
10000000 10000000
*/