摘要：
A: 1分钟题,往后扫一遍int a[MAXN];int vis[MAXN];int main(){ int n,m; cin>>n>>m; MEM(vis,0); for(int i = 1 ; i >a[i]; for(int j = a[i] ; j >n>>a>>b; for(int i = 0 ; i >x[i]; LL p = (x[i]*a)/b; if((p*b)%a == 0){ cout>n>>k){ int m = n/2; if(n == 1 && ... 阅读全文
