【ZOJ】3609 Modular Inverse
1. 题目描述
求乘法逆元。
2. 基本思路
利用扩展gcd求逆元,模板题目。
3. 代码
/* 3609 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <bitset>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000") #define sti set<int>
#define stpii set<pair<int, int> >
#define mpii map<int,int>
#define vi vector<int>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define rep(i, a, n) for (int i=a;i<n;++i)
#define per(i, a, n) for (int i=n-1;i>=a;--i)
#define clr clear
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1 #define LL long long void egcd(LL a, LL b, LL& g, LL& x, LL& y) {
if (!b) {
g = a;
x = ;
y = ;
} else {
egcd(b, a%b, g, y, x);
y -= a/b*x;
}
} int inv(LL a, LL n) {
LL g, x, y; egcd(a, n, g, x, y);
if (g != )
return -; return x%n== ? n:(x%n+n)%n;
} int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif int t;
LL a, m;
LL ans; scanf("%d", &t);
while (t--) {
scanf("%lld%lld", &a, &m);
ans = inv(a, m);
if (ans == -) {
puts ("Not Exist");
} else {
printf("%lld\n", ans);
}
} #ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif return ;
}
4. 代码生成器
import sys
import string
from random import randint def GenData(fileName):
with open(fileName, "w") as fout:
t = 1000
fout.write("%d\n" % (t))
for tt in xrange(t):
n = randint(1, 1000)
m = randint(1, 1000)
fout.write("%d %d\n" % (n, m)) def MovData(srcFileName, desFileName):
with open(srcFileName, "r") as fin:
lines = fin.readlines()
with open(desFileName, "w") as fout:
fout.write("".join(lines)) def CompData():
print "comp"
srcFileName = "F:\Qt_prj\hdoj\data.out"
desFileName = "F:\workspace\cpp_hdoj\data.out"
srcLines = []
desLines = []
with open(srcFileName, "r") as fin:
srcLines = fin.readlines()
with open(desFileName, "r") as fin:
desLines = fin.readlines()
n = min(len(srcLines), len(desLines))-1
for i in xrange(n):
ans2 = int(desLines[i])
ans1 = int(srcLines[i])
if ans1 > ans2:
print "%d: wrong" % i if __name__ == "__main__":
srcFileName = "F:\Qt_prj\hdoj\data.in"
desFileName = "F:\workspace\cpp_hdoj\data.in"
GenData(srcFileName)
MovData(srcFileName, desFileName)
【ZOJ】3609 Modular Inverse的更多相关文章
- zjuoj 3609 Modular Inverse
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3609 Modular Inverse Time Limit: 2 Seco ...
- 【ZOJ】4012 Your Bridge is under Attack
[ZOJ]4012 Your Bridge is under Attack 平面上随机n个点,然后给出m条直线,问直线上有几个点 \(n,m \leq 10^{5}\) 由于共线的点不会太多,于是我们 ...
- 【线性代数】2-5:逆(Inverse)
title: [线性代数]2-5:逆(Inverse) toc: true categories: Mathematic Linear Algebra date: 2017-09-11 20:00:1 ...
- ZOJ 3609 Modular Inverse(拓展欧几里得求最小逆元)
Modular Inverse Time Limit: 2 Seconds Memory Limit: 65536 KB The modular modular multiplicative ...
- ZOJ——3609 Modular Inverse
Modular Inverse Time Limit: 2 Seconds Memory Limit: 65536 KB The modular modular multiplicative ...
- 【ZOJ】【3329】One Person Game
概率DP/数学期望 kuangbin总结题目中的第三道 看来还是没有进入状态啊……都说是DP了……当然是要找[状态之间的转移关系]了…… 本题中dp[i]跟 dp[i-(k1+k2+k3)] 到dp[ ...
- 【有上下界网络流】【ZOJ】2314 Reactor Cooling
[算法]有上下界网络流-无源汇(循环流) [题解]http://www.cnblogs.com/onioncyc/p/6496532.html //未提交 #include<cstdio> ...
- ZOJ 3609 Modular Inverse(扩展欧几里德)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4712 The modular modular multiplicat ...
- ZOJ 3609 Modular Inverse
点我看题目 题意 : 这个题是求逆元的,怎么说呢,题目看着很别扭....就是给你a和m,让你求一个最小的x满足a-1≡x (mod m).或者ax≡1 (mod m).通俗点说呢,就是找一个最小的x, ...
随机推荐
- 解决Ajax跨域问题:Origin xx is not allowed by Access-Control-Allow-Origin.
一:使用jsonp格式, 如jquery中ajax请求参数 dataType:'JSONP'. <html> <head> <title>title</t ...
- Linux系统下ssh的相关配置详细解析
Linux系统下ssh的相关配置进行了详细的分析介绍. ssh是大家常用的登录linux服务器的方式,但是为了安全考虑,有时候我们需要针对ssh做一些特殊处理,本文记录笔者曾经做过的一些修改,供大家参 ...
- NEV_SDK开发环境部署手册
根据项目开发需求,要在MEC服务器上部署如下内容:Nginx.Nginx push stream module.Jason CPP.Spawn-fcgi.libfcgi.Redis.Hiredis.B ...
- 通过telnet命令进行网络邮件发送
1.建立smtp邮箱服务连接 open smtp.sina.com 2.连接上邮箱服务后进行握手操作 helo smtp.sina.com 3.输入帐号密码进行验证::此步后缓冲区会输出一些字符,你只 ...
- 泛形集合List<T>
public class Person { /// <summary> /// 姓名 /// </summary> private string name; public st ...
- EXTJS API
EXTJS API 链接: http://docs.sencha.com/extjs/5.0.0/ http://docs.sencha.com/extjs/4.2.2/ http://docs.se ...
- UUID为36位
package util; import java.util.UUID; public class UUIDUtil { public static UUID getId(){ return UUID ...
- python学习笔记6(字典)
映射:键值对的关系,键(key)映射值(value) 字典是Python唯一的映射类型 >>> phonebook = {'} >>> phonebook {'} ...
- GWT RPC
GWT RPC GWT RPCRemote Procedure Calls GWT: Google Web Toolkit的缩写,有了 GWT可以使用 Java 编程语言编写 AJAX 前端,然后 G ...
- 企业应用的Web程序的安全性
提起安全性这个话题,大家恐怕依稀还记得Sony的PSP账户信息泄露的事故造成的重大损失.但是又隐隐觉得这事儿离我很远,无需过多考虑.也有的人会想,我们做的是企业内部系统所以不必太在意.但是,Web程序 ...