HDU5514 容斥原理

每次遇到容斥，都只会直接上莫比乌斯系数，方便快捷，这次1e9的GG了，实际上还是容斥，只不过这次并不需要很多数的容斥，只需要m的因子之间互相容斥一下就可以了。

参考一份题解很不错，他文中提到的错误解法我们队训练时也这么想的。。http://m.blog.csdn.net/kevin66654/article/details/52724774

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
long long tmp[11234];
long long tmp2[11234];
long long p[112342];
int tt[112342];
long long gcd(long long x,long long y){
    return y==0?x:gcd(y,x%y);
}
int main()
{
    int T;
    scanf("%d",&T);
    for (int cas = 1 ; cas <= T; cas++){
        int n;
        long long m;
        scanf("%d%lld",&n,&m);

        for (int i = 1; i<=n; i++){
            scanf("%lld",&tmp[i]);
            tmp[i] = gcd(tmp[i],m);
        }
        sort(tmp+1,tmp+1+n);
        if (tmp[1]==1){
            printf("Case #%d: %lld
",cas,m*(m-1)/2);
            continue;
        }
        int cnt = 0;
        for (int i = 1 ; i<=n ;i++){
            int flag = 1;
            for (int j = 1; j<i ; j++){
                if (tmp[i] % tmp[j] == 0 ) {
                    flag = 0;
                    break;
                }
            }
            if (flag)tmp2[++cnt] = tmp[i];
        }
        tmp2[cnt+1] = mod;
        int cntt = 0;
        for (int i = 2 ; i*i<=m; i++){
            if (m%i==0) {
                p[++cntt] = i;
                if (i*i!=m) p[++cntt] = m/i;
            }
        }
        memset(tt,0,sizeof(tt));
        sort(p+1,p+cntt+1);
        int pos = 1;
        long long ans = 0;
        for (int i = 1; i<=cntt ; i++){
            if (tmp2[pos] == p[i]) ans += (m/tmp2[pos])*(m/tmp2[pos]-1)/2*tmp2[pos],pos++,tt[i] = 1;
            else{
                int flag = 1;
                for (int j = 1; tmp2[j] < p[i] ; j++){
                    if (p[i] % tmp2[j]==0) {flag = 0;break;}
                }
                if (flag) continue;
                int t = 0;
                for (int j = 1; j < i ; j++){
                    if (p[i] % p[j] == 0) {t+=tt[j];}
                }
                ans += (1-t) * (m/p[i])*(m/p[i]-1)/2*p[i] , tt[i]=(1-t);
            }
        }
        printf("Case #%d: %lld
",cas,ans);
    }
    return 0;
}