HDU 5514 Frogs (数论容斥)

题意：有n只青蛙，m个石头(围成圆圈)。第i只青蛙每次只能条ai个石头，问最后所有青蛙跳过的石头的下标总和是多少？

析：首先可以知道的是第 i 只青蛙可以跳到 k * gcd(ai, m)，然后我就计算所有的等差数列，但是好像如果全算，那么就可能会有重复，所以我们考虑用容斥原理。

先把 m 的所有因数都求出来，然后把 gcd(ai, m)，都标记一下，然后再去计算，多了就减去，少了就加。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> v;
int f[maxn], num[maxn];

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d", &n, &m);
        v.clear();
        for(int i = 1; i*i <= m; ++i) if(m % i == 0){
            v.push_back(i);
            if(i*i != m && i != 1)  v.push_back(m/i);
        }
        sort(v.begin(), v.end());
        memset(num, 0, sizeof num);
        memset(f, 0, sizeof f);
        int x;
        for(int i = 0; i < n; ++i){
            scanf("%d", &x);
            x = gcd(x, m);
            for(int j = 0; j < v.size(); ++j) if(v[j] % x == 0){
                f[j] = 1;
            }
        }
        LL ans = 0;
        for(int i = 0; i < v.size(); ++i) if(f[i] != num[i]){
            int tmp = m / v[i] - 1;
            ans += (LL)m * tmp / 2 * (f[i] - num[i]);
            tmp = f[i] - num[i];
            for(int j = 0; j < v.size(); ++j) if(v[j] % v[i] == 0){
                num[j] += tmp;
            }
        }
        printf("Case #%d: %I64d\n", kase, ans);
    }
    return 0;
}