51nod1237 最大公約數之和
阿新 • • 發佈:2019-03-30
odi cto for inline 51nod tor main font cpp
題目鏈接
題意
其實就是求
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^ngcd(i,j)\]
思路
建議先看一下此題的一個弱化版
推一下式子
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^ngcd(i,j)\]
\[= \sum\limits_{k=1}^nk\sum\limits_{i=1}^n\sum\limits_{j=1}^n[gcd(i,j)=k]\]
\[=\sum\limits_{k=1}^nk\sum\limits_{i=1}^{\frac{n}{k}}\sum\limits_{j=1}^{\frac{n}{k}}[gcd(i,j)=1]\]
\[=\sum\limits_{k=1}^nk\sum\limits_{i=1}^{\frac{n}{k}}2\varphi(i)-1\]
設\(\phi(i)=\varphi(1)+\varphi(2)+...+\varphi(i)\)
則原式
\[=\sum\limits_{i=1}^ni(2\phi(\frac{n}{i})-1)\]
然後就可以數論分塊啦。
至於怎麽比較快的求\(\phi(i)\),基本的杜教篩嘍。。
代碼
//loj6074 /* * @Author: wxyww * @Date: 2019-03-30 12:43:48 * @Last Modified time: 2019-03-30 19:43:10 */ #include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<algorithm> #include<queue> #include<map> #include<vector> #include<ctime> using namespace std; typedef long long ll; const int mod = 1e9 + 7,N = 1000000 + 100,inv2 = (mod + 1) / 2; ll read() { ll x=0,f=1;char c=getchar(); while(c<'0'||c>'9') { if(c=='-') f=-1; c=getchar(); } while(c>='0'&&c<='9') { x=x*10+c-'0'; c=getchar(); } return x*f; } map<ll,ll>ma; ll n,sum[N]; int dis[N],vis[N],js; int dls(ll x) { if(x <= 1000000) return sum[x]; if(ma[x]) return ma[x]; ll ret = 1ll * x % mod * ((x + 1) % mod) % mod * inv2 % mod; for(ll l = 2,r;l <= x;l = r + 1) { r = x / (x / l); ret -= 1ll * (r - l + 1) % mod * dls(x / l) % mod; ret = (ret + mod) % mod; } return ma[x] = ret; } void pre() { sum[1] = 1;vis[1] = 1; int NN = min(n,1000000ll); for(int i = 2;i <= NN;++i) { if(!vis[i]) { dis[++js] = i; sum[i] = i - 1; } for(int j = 1;j <= js && dis[j] * i <= NN;++j) { vis[dis[j] * i] = 1; if(i % dis[j] == 0) { sum[dis[j] * i] = sum[i] * dis[j] % mod; break; } sum[dis[j] * i] = (dis[j] - 1) * sum[i] % mod; } sum[i] += sum[i - 1]; sum[i] %= mod; } } signed main() { n = read(); pre(); ll ans = 0; for(ll l = 1,r;l <= n;l = r + 1) { r = n / (n / l); ans = (ans + (1ll * (r - l + 1) % mod * ((r + l) % mod) % mod * inv2 % mod) % mod * ((2ll * dls(n / l) % mod) - 1 + mod) % mod) % mod; } cout<<ans; return 0; }
51nod1237 最大公約數之和