bzoj 2190: [SDOI2008]仪仗队 【容斥】

题目链接：bzoj 2190: [SDOI2008]仪仗队

去掉第一行和第一列，然后容斥。之后加2就好了。

$AC$代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e6+1;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void getmax(int &a, int b) {a = max(a, b); }
void getmin(int &a, int b) {a = min(a, b); }
void add(LL &x, LL y) { x += y; x %= MOD; }
int p[1001], k;
void getp(int n)
{
    k = 0;
    for(int i = 2; i * i <= n; i++)
    {
        if(n % i == 0) {
            p[k++] = i;
            while(n % i == 0) n /= i;
        }
    }
    if(n > 1) p[k++] = n;
}
LL q[MAXN];
LL nop(LL m)
{
    int top = 0, t;
    q[top++] = -1;
    for(int i = 0; i < k; i++)
    {
        t = top;
        for(int j = 0; j < t; j++) {
            q[top++] = q[j] * p[i] * (-1);
        }
    }
    LL sum = 0;
    for(int i = 1; i < top; i++)
        sum += m / q[i];
    return sum;
}
int main()
{
    int n; scanf("%d", &n); n--;
    int ans = n;
    for(int i = 2; i <= n; i++)
    {
        getp(i);
        ans += n - nop(n);
    }
    cout << ans + 2 << endl;
    return 0;
}