随笔分类 -  容斥原理

摘要:Problem Description Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.Two inte 阅读全文
posted @ 2017-07-30 20:55 ⊙∽⊙Perseverance 阅读(117) 评论(0) 推荐(0)
摘要:1 #include 2 int num[6] = {2,3,5,7}; 3 int n; 4 int solve() 5 { 6 int ans = 0; //是那四个数的倍数的数的数量 7 for (int i = 1 ; i < (1<<4) ; i++) //选数 8 { 9 int ant = 0... 阅读全文
posted @ 2017-07-30 18:03 ⊙∽⊙Perseverance 阅读(184) 评论(0) 推荐(0)