数论——bzoj3560: DZY Loves Math V

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http://www.lydsy.com/JudgeOnline/problem.php?id=3560
我要好好练习数论了，今日见到数论一点思路都没有，唉；
公示推倒看popoqqq的好了，写的贼好；
http://blog.csdn.net/popoqqq/article/details/42739963
当然队欧拉函数的理解是基础；
代码

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#define Ll long longusing namespace std;struct cs{int x,y;}q[1000000];int a[100005];int n,m,top,x,last,mo=1e9+7;Ll f[10000],ans=1;void make(int x){    for(int i=2;i*i<=x;i++)        if(x%i==0){            q[++top].x=i;            for(;x%i==0;x/=i)q[top].y++;        }    if(x!=1)q[++top].x=x,q[top].y=1;}bool cmp(cs a,cs b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;}Ll ksm(Ll x,Ll y){    Ll ans=1;    for(;y;y>>=1,x=x*x%mo)if(y&1)ans=ans*x%mo;    return ans;}Ll find(int x,int y){    f[0]=1; Ll temp=1;    for(int i=1;i<=q[y].y;i++)f[i]=f[i-1]*q[x].x%mo;    for(int i=1;i<=q[y].y;i++)f[i]=(f[i]+f[i-1])%mo;    for(int i=x;i<=y;i++)temp=temp*f[q[i].y]%mo;    temp=(temp-1)*ksm(q[x].x,mo-2)%mo*(q[x].x-1)+1;    return temp%mo;}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)scanf("%d",&x),make(x);    sort(q+1,q+top+1,cmp);    for(int i=1;i<=top;i++)        if(q[i].x!=q[i+1].x){            ans=ans*find(last+1,i)%mo;            last=i;        }    printf("%lld",ans);}

然后我的同学还有一种写法，就是吧上面的公示化开用递推；我不会啊

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>#include<string>#include<map>#include<cstring>#include<ctime>#include<vector>#define inf 1e9#define ll long long#define For(i,j,k) for(ll i=j;i<=k;i++)#define Dow(i,j,k) for(ll i=k;i>=j;i--)using namespace std;int n;ll ans=1,zy[1000001],cnt[1000001],tot;int tim[1000001],tmp[1000001];ll mo=1e9+7;void get(int &x){    char c = getchar(); x = 0;    while(c < '0' || c > '9') c = getchar();    while(c <= '9' && c >= '0') x = x*10+c-48, c = getchar();}int main(){    freopen("phi.in","r",stdin);    freopen("phi.out","w",stdout);    get(n);    For(i,1,1000000)    zy[i]=tim[i]=tmp[i]=1;      For(i,1,n)    {        int x;        get(x);        int t=sqrt(x);        for (int y=2;y*y<=x;y++)        {            while(x%y==0)            {                if(tim[y]!=i)   tim[y]=i,tmp[y]=zy[y];  //cout<<zy[y]*y<<' '<<tmp[y]-1<<' '<<y<<endl;                zy[y]=(zy[y]*y+tmp[y]-1)%mo;                x/=y;               }           }        tim[x]=i,tmp[x]=zy[x];        zy[x]=(zy[x]*x+tmp[x]-1)%mo;    }    For(i,1,1000000)    ans=(ans*zy[i])%mo;    printf("%d",ans);}

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