ACM学习历程—广东工业大学2016校赛决赛－网络赛F 我是好人4(数论)

题目链接：http://gdutcode.sinaapp.com/problem.php?cid=1031&pid=5

这个题目一看就是一道数论题，应该考虑使用容斥原理，这里对lcm进行容斥。

不过直接上去是T，考虑到序列中同时存在i和ki的话，其实只需要考虑i，所以先对序列中为倍数的对进行处理。

这里的容斥用了hqw的写法。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <string>
#define LL long long

using namespace std;

const LL maxN = 1e9;
int n;
int t[60], a[60], len;
LL ans;

void input()
{
scanf("%d", &n);
len = n;
for (int i = 0; i < len; ++i)
scanf("%d", &t[i]);
for (int i = 0; i < len; ++i)
{
if (t[i] == -1) continue;
for (int j = 0; j < len; ++j)
{
if (i == j) continue;
if (t[j] == -1) continue;
if (t[j]%t[i] == 0) t[j] = -1;
}
}
int top = 0;
for (int i = 0; i < len; ++i)
if (t[i] != -1)
a[top++] = t[i];
len = top;
sort(a, a+len);
}

LL gcd(LL x, LL y)
{
LL r;
while (y != 0)
{
r = y;
y = x%y;
x = r;
}
return x;
}

LL lcm(LL x, LL y)
{
return x/gcd(x, y)*y;
}

void dfs(int now, LL num, int sz)
{
if (now == len)
{
if (sz)
{
LL last = maxN/num;
if (sz&1) ans -= last;
else ans += last;
}
return;
}
if (num%a[now] == 0) return;
dfs(now+1, num, sz);
LL t = lcm(num, a[now]);
if (t <= maxN) dfs(now+1, t, sz+1);
}

void work()
{
if (len == 1 && a[0] == 1) printf("0\n");
else
{
ans = maxN;
dfs(0, 1, 0);
cout << ans << endl;
}
}

int main()
{
//freopen("test.in", "r", stdin);
int T;
scanf("%d", &T);
for (int times = 1; times <= T; ++times)
{
input();
work();
}
return 0;
}

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