hdu 4944

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题意：for(int i=1;i<=n;i++)
for(int j = 1;j <= n; j++)
for(int k = 1;k <= n; k++)
if(i % k == 0 && j % k == 0) ans += i * j / gcd(i, j);

易得dp[i] = dp[i-1] + num[i] * i;
关于num[i]的求法：
如num[4]
k == 1: 1/gcd(1,4) 2/gcd(2,4) 3/gcd(3,4) 4/gcd(4,4)
k == 2: 2/gcd(1,2) 4/gcd(2,2)
k == 4: 4/gcd(1,1)
整理得：1/1
2/1 2/2
3/1
4/1 4/2 4/4
易得分母为i的因子，分子为因子的倍数。
竖着看，分母相同的呈等差数列。所以num[i]可以logn求的
代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <functional>#include <sstream>#include <iomanip>#include <cmath>#include <cstdlib>#include <ctime>//#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;#define INF 1e9#define MAXN 21#define maxn 500000const ll mod = 1LL<<32;#define eps 1e-7#define pi 3.1415926535897932384626433#define rep(i,n) for(int i=0;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define scan(n) scanf("%d",&n)#define scan2(n,m) scanf("%d%d",&n,&m)#define scans(s) scanf("%s",s);#define ini(a) memset(a,0,sizeof(a))#define out(n) printf("%d\n",n)ll gcd(ll a,ll b) {return b==0?a:gcd(b,a%b);}using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1ll num[maxn+1], dp[maxn+1];void init(){num[0] = 0;for(ll i = 1;i <= maxn; i++){for(ll j = i; j <= maxn; j += i){num[j] += (j/i+1) * (j/i) / 2;if(num[j] >= mod) num[j] %= mod;}}for(int i = 1;i <= maxn; i++){dp[i] = dp[i-1] + i * num[i];if(dp[i] >= mod) dp[i] %= mod;}}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);  //  freopen("out.txt","w",stdout);#endifinit();int T;cin>>T;int cas = 1;int n;while(T--){scan(n);printf("Case #%d: %I64d\n",cas++, dp[n]);}return 0;}

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