九月旧约 - 博客园

牛客多校第五场 E room 二分图匹配 KM算法模板

摘要：链接：https://www.nowcoder.com/acm/contest/143/E来源：牛客网 Nowcoder University has 4n students and n dormitories ( Four students per dormitory). Students num 阅读全文

posted @ 2018-08-03 10:01 九月旧约阅读(266) 评论(0) 推荐(0)

牛客第五场多校 J plan 思维

摘要：链接：https://www.nowcoder.com/acm/contest/143/J来源：牛客网 There are n students going to travel. And hotel has two types room:double room and triple room. Th 阅读全文

posted @ 2018-08-02 21:05 九月旧约阅读(296) 评论(0) 推荐(0)

牛客第五场 G max 思维

摘要：链接：https://www.nowcoder.com/acm/contest/143/G来源：牛客网 Give two positive integer c, n. You need to find a pair of integer (a,b) satisfy 1<=a,b<=n and the 阅读全文

posted @ 2018-08-02 20:49 九月旧约阅读(152) 评论(0) 推荐(0)

牛客第五场多校 A gpa 分数规划（模板）

摘要：链接：https://www.nowcoder.com/acm/contest/143/A来源：牛客网 Kanade selected n courses in the university. The academic credit of the i-th course is s[i] and th 阅读全文

posted @ 2018-08-02 20:35 九月旧约阅读(302) 评论(0) 推荐(0)

杭电多校第四场 Problem K. Expression in Memories 思维模拟

摘要： Problem K. Expression in Memories Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0 Accepted Su 阅读全文

posted @ 2018-08-01 16:28 九月旧约阅读(273) 评论(0) 推荐(0)

HDU5461 Largest Point 思维 2015沈阳icpc

摘要： Largest Point Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3065 Accepted Submission(s): 1078 P 阅读全文

posted @ 2018-07-31 21:55 九月旧约阅读(431) 评论(0) 推荐(0)

HDU 5451 Best Solver 数论快速幂 2015沈阳icpc

摘要： Best Solver Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 1115 Accepted Submission(s): 645 Pro 阅读全文

posted @ 2018-07-31 21:45 九月旧约阅读(153) 评论(0) 推荐(0)

HDU 4565 So Easy! 广义斐波拉数数论 (a+sqrt(b))^n%mod 模板

摘要： So Easy! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5905 Accepted Submission(s): 1966 Proble 阅读全文

posted @ 2018-07-31 21:24 九月旧约阅读(230) 评论(0) 推荐(0)

摘要： #include #include #include #include #include #include #include #include #include using namespace std; #define rep(i,a,n) for (int i=a;i=a;i--) #define pb push_back #define mp make_pair #defi... 阅读全文

posted @ 2018-07-31 18:53 九月旧约阅读(161) 评论(0) 推荐(0)

牛客小白赛5 无关(relationship) 容斥原理（计算因子数的模板）

摘要：链接：https://www.nowcoder.com/acm/contest/135/A来源：牛客网若一个集合A内所有的元素都不是正整数N的因数，则称N与集合A无关。给出一个含有k个元素的集合A={a1,a2,a3,...,ak}，求区间[L,R]内与A无关的正整数的个数。保证A内的元素都是阅读全文

posted @ 2018-07-30 10:41 九月旧约阅读(212) 评论(0) 推荐(0)