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摘要: ll random(ll n) { return (ll)((double)rand()/RAND_MAX*n + 0.5); } ll pow_mod(ll a,ll p,ll n) { if(p == 0) return 1; ll ans = pow_mod(a,p/2,n); ans = ans*ans%n; if(p%2) ... 阅读全文
posted @ 2014-06-30 21:45 whatbeg 阅读(305) 评论(0) 推荐(0) 编辑