H_Fighter-优快云博客

原创 POJ3233 Matrix Power Series

传送门：http://poj.org/problem?id=3233 代码如下： #include #include #include using namespace std; int k,n,m; struct matrix { int m[31][31]; void init() { memset(m,0,sizeof(m));

2017-04-22 20:14:12 308

原创 HDU2689 Sort it（树状数组求逆序数）

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2689 代码如下：#include #include #include #define maxn 10000 using namespace std; int n,tree[maxn]; int lowbit(int i) { return i&(-i); } int update(int

2017-04-09 09:40:08 461

原创 POJ3970 Party

Party Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1300 Accepted: 498 Description The CEO of ACM (Association of Cryptographic Mavericks) organization h

2017-04-01 19:11:01 607

原创 POJ2142 The Balance（扩展欧几里得算法）

The Balance Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 6473 Accepted: 2849 Description Ms. Iyo Kiffa-Australis has a balance and only two kinds of wei

2017-04-01 11:57:43 732

原创 POJ2115 C Looooops（扩展欧几里得算法）

C Looooops Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 25936 Accepted: 7346 Description A Compiler Mystery: We are given a C-language style for loop of

2017-04-01 11:02:29 373

原创 POJ1061 青蛙的约会（扩展欧几里得）

青蛙的约会 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 113074 Accepted: 23054 Description 两只青蛙在网上相识了，它们聊得很开心，于是觉得很有必要见一面。它们很高兴地发现它们住在同一条纬度线上，于是它们约定各自朝西跳，直到碰

2017-04-01 09:25:05 518

原创 HDU3074 Multiply game（线段树单点更新）

Multiply game Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2540 Accepted Submission(s): 894 Problem Description Tired of play

2017-03-30 16:45:09 291

原创 POJ2488 A Knight's Journey(深搜)

A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 44467 Accepted: 15107 Description Background The knight is getting bored of seeing the

2017-03-27 20:51:39 443

原创 POJ2478 Farey Sequence

Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are F2

2017-03-12 20:11:21 357

原创 POJ1284 Primitive Roots（原根定理）

Primitive Roots Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is equal to { 1, …, p-1 }. For example, the

2017-03-12 20:02:37 571

原创欧拉函数的算法实现

欧拉函数：在数论，对正整数n，欧拉函数是小于等于n的数中与n互质的数的数目。公式：∅(x)=(x)(1-1/p1)(1-1/p2)······(1-1/pn)，其中p1···pn是x的素因数。long long phi(long long int n） { long long res=n; for(long long i=2;i*i<=n;i++) {//任何一

2017-03-12 19:25:36 880

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