HDU4059

hdu4059 The Boss on Mars(差分+容斥原理)

题意：求小于n(1≤n≤10^8)的数中，与n互质的数的四次方和。 知识点：   差分：        一阶差分：设                         则  为一阶差分。        二阶差分：                                                                                      n阶差分：   且
strangedbly·2016-04-10 21:00

hdu4059 The Boss on Mars

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1335    Accepted Submission(s): 401
·2015-11-13 12:43

hdu4059容斥原理

求1-n中与n互质的数的4次方之和importjava.io.BufferedInputStream; importjava.io.BufferedReader; importjava.io.IOException; importjava.io.InputStream; importjava.io.InputStreamReader; importjava.io.PrintWriter; impo
u013491262·2015-08-27 13:00

hdu4059 The Boss on Mars

TheBossonMarsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1335    AcceptedSubmission(s):401ProblemDescriptionOnMars,thereisahugecompanycalledACM(AhugeC
u010422038·2013-09-02 21:00

hdu4059 容斥原理的好题

1.首先要知道4次方求和的公式2.注意求和公式中有除以30应该乘以30的逆3.容斥原理即原来的所有和- 1个质因数的倍数+2个质因数的倍数-3个质因数的倍数+4个质因数的倍数……每个质因数的贡献就是getsum(n/link[i])*four(link[i])，比如质因数是2时，就是(1^4+2^4+3^4+4^4......+(n/2)^4)*(2^4)还要减掉dfs(i+1,n/link[i]
Jackyguo1992·2013-07-04 16:00
上一页 1 下一页
按字母分类： ABCDEFGHIJKLMNOPQRSTUVWXYZ其他