HDU 2802 F(N)（简单题，找循环解）

【HDU 2802 F(N)（简单题，找循环解）】的更多相关文章

HDU 2802 F(N)（简单题，找循环解）

题目链接 F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4579 Accepted Submission(s): 1610 Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a si…

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超级无敌简单题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 570 Accepted Submission(s): 274 Problem Description 通常来说,题面短的题目一般都比较难,所以我要把题面写得很长很长.通常来说,题面短的题目一般都比较难,所以我要把题面写得很长很长.通常来说,题面短的题目一般都比较难,所…

HDU - 2802 F(N) （周期）

题目链接:HDU 2009-4 Programming Contest 分析:具有一定的周期性——4018处理下就可以A了 Sample Input Sample Output AC代码: #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<iostream> #include<queue> #include<map>…

数学--数论--HDU 2802 F(N) 公式推导或矩阵快速幂

Giving the N, can you tell me the answer of F(N)? Input Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed. Output For each test case, output on a line the v…

HDU 2802 F(N) 数论+打表

题目大意:f[n]-n^3=f[n-2]-(n-1)^3 (n >=3),f[1]=1,f[2]=7,求f[n]. 题目思路:将n^3移到到等式右边化简的到:f[n]=f[n-2]+3n*(n-1)+1: 因为第n项与第n-2项有关,所以知道了f[1]与f[2]的值可以分奇偶打下表,找到循环节为4018. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #incl…

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Problem Description The school set up three elective courses, assuming that these courses are A, B, C. N classes of students enrolled in these courses.Now the school wants to count the number of students who enrolled in at least one course in each cl…