ftx456789的博客

RXD and mathTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1276    Accepted Submission(s): 697Problem DescriptionRXD is a goo

#include <bits/stdc++.h>using namespace std;long long n=1001733993063167141;long long d=212353;long long c=20190324;long long p=891234941;long long q=1123984201;long long e=82381609393152...

简单数学题Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 703 Accepted Submission(s): 289Problem Description已知F(n)=∑i=1n(i×∑j=inCji)F(n)=\sum...

PowModTime Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1599 Accepted Submission(s): 559Problem DescriptionDeclare:k=∑i=1mφ(i∗n) mo...

链接：https://ac.nowcoder.com/acm/contest/881/B来源：牛客网题意有∫0∞11+x2dx=π2\int_0^\infty \frac{1}{1+x^2}dx=\frac{\pi}{2}∫0∞​1+x21​dx=2π​那么给你一个长度为nnn的序列aia_iai​求解1π∫0∞1∏i=1n(ai2+x2)dx\frac{1}{\pi}\int_0^...

题意已知x0,x1,a,b(1≤109)x_0,x_1,a,b(1\leq10^9)x0​,x1​,a,b(1≤109)，求xn=axn−1+bxn−2( mod m)x_n=ax_{n-1}+bx_{n-2}(\ mod \ m)xn​=axn−1​+bxn−2​( mod m)1≤n≤10106,109&lt;m≤2×1091\leq n\...

题意{f0=0f1=1fn=fn−1+fn−2,n≥2\left\{\begin{matrix}f_0=0\\ f_1=1\\ f_n=f_{n-1}+f_{n-2},n\geq 2\end{matrix}\right.⎩⎨⎧​f0​=0f1​=1fn​=fn−1​+fn−2​,n≥2​给你nnn和mmm求∑i=0nfim %1000000000\sum_{i=0}^{n}...