hdu 5212 Code

题目链接

分析：这道题可以尝试容斥原理或者莫比乌斯反演，其实本质上是一样的。

处理一下每个素数在给定的条件中有多少个倍数，然后容斥原理或者莫比乌斯反演统计一下gcd(x,y)=t的(x,y)的对数，最后统计一下答案，取模。

代码：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)
#define   DEBUG         freopen("d.txt", "r", stdin)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   ll            long long
#define   ull           unsigned long long
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)

const int    INF    = 0x3f3f3f3f;
const ll     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const ll     mod    = 10007;
const int    maxmat = 10;
const ull    BASE   = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int maxn = 1e4 + 5;

std::vector<int> prime;
int u[maxn];
bool mark[maxn];
int cnt[maxn];
int N;

void init() {
memset(mark, false, sizeof(mark));
u[1] = 1;
for (int i = 2; i <= 1e4; i ++) {
if (!mark[i]) {
prime.push_back(i);
u[i] = -1;
}
for
4000
(unsigned j = 0; j < prime.size(); j ++) {
int tmp = prime[j] * i;
if (tmp > 1e4) break;
mark[tmp] = true;
if (i % prime[j] == 0) {
u[tmp] = 0;
break;
}
else
u[tmp] = -u[i];
}
}
}

void get_c(int x) {
for (int i = 2; i * i <= x; i ++) {
if (x % i == 0) {
cnt[i] ++;
if (i * i != x) cnt[x / i] ++;
}
}
cnt[x] ++;
}

int main(int argc, char const *argv[]) {
// DEBUG;
init();
while (~scanf("%d", &N)) {
memset(cnt, 0, sizeof(cnt));

ll ans = 0;
int m = -1;
for (int i = 0; i < N; i ++) {
int x; scanf("%d", &x);
ans += (ll)(x - 1) * x % mod;
m = max(x, m);
get_c(x);
}

for (int i = 2; i <= m; i ++) {
ll num = 0;
for (int j = i; j <= m; j += i)
num += (ll)u[j / i] * cnt[j] * (cnt[j] - 1)；
ans = (ans + num * (ll)(i - 1) * i % mod) % mod;
}

printf("%I64d\n", ans);
}
return 0;
}