bzoj 3512 DZY Loves Math IV

xiaoxiao2021-02-28  100

3512: DZY Loves Math IV Time Limit: 15 Sec Memory Limit: 128 MB Submit: 442 Solved: 219 [Submit][Status][Discuss] Description

给定n,m,求 模10^9+7的值。

Input

仅一行,两个整数n,m。

Output

仅一行答案。

Sample Input

100000 1000000000

Sample Output

857275582

数据规模:

1<=n<=10^5,1<=m<=10^9,本题共4组数据。

HINT

Source

By Jc 鸣谢杜教


【分析】 公式恐惧症 还有…DZY是谁


【代码】

//IV #include<map> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define N 100000 #define ll long long #define M(a) memset(a,0,sizeof a) #define fo(i,j,k) for(i=j;i<=k;i++) using namespace std; const int mod=1e9+7; const int mxn=100005; ll ans; int n,m; bool vis[mxn]; map <ll,int> A,B; int k[mxn],phi[mxn],pri[mxn],sum[mxn]; inline void init() { int i,j; phi[1]=k[1]=1; fo(i,2,N) { if(!vis[i]) pri[++pri[0]]=i,phi[i]=i-1,k[i]=1; for(j=1;j<=pri[0] && i*pri[j]<=N;j++) { vis[i*pri[j]]=1; if(i%pri[j]==0) { phi[i*pri[j]]=phi[i]*pri[j]; k[i*pri[j]]=k[i]*pri[j]; break; } k[i*pri[j]]=k[i]; phi[i*pri[j]]=phi[i]*(pri[j]-1); } } fo(i,1,N) sum[i]=(phi[i]+sum[i-1])%mod; } inline int P(int n) { if(n<=N) return sum[n]; if(B.count(n)) return B[n]; ll res=(ll)n*(n+1)/2%mod; for(int i=2,last=0;i<=n;i=last+1) { last=n/(n/i); res=(res-(ll)(last-i+1)*P(n/i))%mod; } return B[n]=(res+mod)%mod; } inline ll S(int n,int m) { ll res=0; if(!m) return 0; if(A.count((ll)n*mod+m)) return A[(ll)n*mod+m]; if(n==1) return A[(ll)n*mod+m]=P(m); for(int i=1;i*i<=n;i++) if(n%i==0) { res=(res+(ll)phi[n/i]*S(i,m/i)%mod)%mod; if(i*i!=n) res=(res+(ll)phi[i]*S(n/i,m/(n/i))%mod)%mod; } return A[(ll)n*mod+m]=res; } int main() { init(); int i,j; scanf("%d%d",&n,&m); fo(i,1,n) ans=(ans+(ll)k[i]*S(i/k[i],m)%mod)%mod; printf("%lld\n",ans); return 0; }
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