HDU5528 - Count a * b

HDU5528 - Count a * b

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做法：求\(\sum_{m|n}(m^2 - \sum_{i=1}^{m}\sum_{j=1}^m [m|(ij)])\)

\(h(m) = \sum_{i=1}^{m}\sum_{j=1}^m [m|(ij)] = \sum_{i=1}^m\sum_{j=1}^m [\frac{m}{(i,m)}|\frac{i}{(i,m)}j]\)
$ = \sum_{i=1}^m\sum_{j=1}^m [\frac{m}{(i,m)}|j] = \sum_{i=1}^m \frac{m}{\frac{m}{(i,m)}} = \sum_{i=1}^m (i,m)$
$ = \sum_{d|m} d \sum_{i=1}^m [(i,m)=d] = \sum_{d|m} d \sum_{i=1}^{\frac{m}{d}} [(i,\frac{m}{d})=1] = \sum_{d|m}d\varphi(\frac{m}{d})$

$
\sum_{m|n} \sum_{d|m}d\varphi(\frac{m}{d}) = \sum_{d|n}d\sum_{m|\frac{n}{d}}\varphi(m) = \sum_{d|n}d\frac{n}{d} = \sum_{d|n}n
$
所以\(ans = \sum_{d|n} d^2 - n\sigma_0(n)\)两个值均为积性函数，预处理素因子，合并答案即可。本题时限较紧，不能直接使用试除法分解，尽量优化常数。

#include <bits/stdc++.h>
typedef unsigned long long ll;
const int N = 1e6 + 7;
inline int read() {char c = getchar(); int x = 0, f = 1;while( ! isdigit(c) ) { if(c == '-') f = -1; c = getchar(); }while( isdigit(c) ) { x = x*10 + c - '0'; c = getchar(); }return x * f;
}
inline void write( ll x ) {if( x >= 10 ) write( x / 10 );putchar( x % 10 + '0' );
}
using namespace std;
int T;
ll n;
int p[N], notp[N];
void init() {notp[1] = 1;for(int i = 2; i <= 1e6; ++i) {if(!notp[i]) p[++p[0]] = i;for(int j = 1; j <= p[0] && p[j]*i <= 1e6; ++j) {notp[i*p[j]] = 1;if(i%p[j] == 0) break;}}
}
ll solve(ll n) {ll ans = 1, num = 1, tn = n;for(int i = 1; i <= p[0] && p[i]*p[i] <= n; ++i) if(n%p[i] == 0){ll t = p[i], tmp1 = 1, tmp2 = 0;while(n%t==0) {n/=t;tmp1 = tmp1*t*t + 1;++tmp2;}ans *= tmp1;num *= (tmp2 + 1);}if(n!=1) {num*=2;ans *= (1 + n*n);}ans -= num*tn;return ans;
}
int main() {T = read();init();while(T--) {n = read();write(solve(n)); putchar('\n');}return 0;
}