Codeforces Round #284 (Div. 1) C. Array and Operations 二分图匹配

因为只有奇偶之间有操作， 可以看出是二分图， 然后拆质因子， 二分图最大匹配求答案就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;

int odd[N], even[N], ocnt, ecnt;
int L[N], R[N];
int G[N][N];
int match[N];
bool vis[N];
int n, m;

int path(int u) {
    for(int v = ; v <= ecnt; v++) {
        if(G[u][v] && !vis[v]) {
            vis[v] = true;
            if(match[v] == - || path(match[v])) {
                match[v] = u;
                return ;
            }
        }
    }
    return ;
}

int main() {
    memset(match, -, sizeof(match));
    scanf("%d%d", &n, &m);
    for(int i = ; i <= n; i++) {
        int x;
        scanf("%d", &x);
        if(i & ) {
            L[i] = ocnt + ;
            for(int j = ; j * j <= x; j++) {
                if(x % j) continue;
                while(x % j == ) {
                    odd[++ocnt] = j;
                    x /= j;
                }
            }
            if(x > ) odd[++ocnt] = x;
            R[i] = ocnt;
        } else {
            L[i] = ecnt + ;
            for(int j = ; j * j <= x; j++) {
                if(x % j) continue;
                while(x % j == ) {
                    even[++ecnt] = j;
                    x /= j;
                }
            }
            if(x > ) even[++ecnt] = x;
            R[i] = ecnt;
        }
    }
    while(m--) {
        int u, v; scanf("%d%d", &u, &v);
        if(v & ) swap(u, v);
        for(int i = L[u]; i <= R[u]; i++)
            for(int j = L[v]; j <= R[v]; j++)
                if(odd[i] == even[j]) G[i][j] = ;
    }
    int ans = ;
    for(int i = ; i <= ocnt; i++) {
        memset(vis, false, sizeof(vis));
        if(path(i)) ans++;
    }
    printf("%d\n", ans);
    return ;
}

/*
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