Luogu3455:莫比乌斯反演进行GCD计数 莫比乌斯反演就是用来解决这一类问题的,通常f函数是要求的那个,F函数是显然的 这样利用F的结果就可以推出来f的结果 在计算结果的时候整除分快儿一下就可以很快了 #include<cstdio> #include<algorithm> using std::min; ; int cnt; long long ans; bool vis[maxn]; int mu[maxn],sum[maxn]; long long prim[maxn]
GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4291 Accepted Submission(s): 1502 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)
GCD SUM Time Limit: 8000/4000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description 给出N,M执行如下程序:long long ans = 0,ansx = 0,ansy = 0;for(int i = 1; i <= N; i ++) for(int j = 1; j <= M; j ++)