Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5221    Accepted Submission(s): 1761 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10).   Input 输入数据的第一行为一个正整数…

Strange Way to Express Integers Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following: Choose k different positive integers a1, a2, …, ak. For some…

不互质的中国剩余定理…… 链接http://acm.hdu.edu.cn/showproblem.php?pid=3579 #include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<iomanip>;}…

Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4206    Accepted Submission(s): 1616 Problem Description One day I was shopping in the supermarket. There was a cashier counting coins…

好久没写什么数论,同余之类的东西了. 昨天第一次用了剩余定理解题,今天上百度搜了一下hdu中国剩余定理.于是就发现了这个题目. 题目的意思很简单.就是告诉你n个m[i],和n个a[i].表示一个数对m[i]取模的值为a[i]. 乍一看以为这个题目可以用中国剩余定理,但是看仔细了吗?这里的mi并没有说是互素的哦. 肿么办?只能用安叔以前说的解同余方程的那种土方法了. 上代码:(我稍微改进了一下,根据题目的具体情况). #include <iostream> #include <cstdio…

http://acm.hdu.edu.cn/showproblem.php?pid=5072 求n个不同的数(<=1e5)中有多少组三元组(a, b, c)两两不互质或者两两互质. 逆向求解,把所有不符合的情况求出来用总的情况数减去即可: 先用容斥求出和a[i] 互质的个数num ,然后不符合条件的 就是 num*(n-1-num); 求法见http://blog.csdn.net/u012774187/article/details/40399567 #include <cstdio>…

<题目链接> 题目大意: 给你一些模数和余数,让你求出满足这些要求的最小的数的值. 解题分析: 中国剩余定理(模数不一定互质)模板题 #include<stdio.h> using namespace std; #define ll long long ll A[],B[];//B[i]为余数 ll dg,ans;//dg为A[i]的最小公倍数 ans 为最小解 void exgcd(ll a, ll b, ll &d, ll&x, ll &y) { ; y…

分析:考虑对给定的出圈序列进行一次模拟,对于出圈的人我们显然可以由位置,编号等关系得到一个同余方程 一圈做下来我们就得到了n个同余方程 对每个方程用扩展欧几里得求解,最后找到最小可行解就是答案. 当然不要忘了判无解的情况. 有非常多选手似乎都是一眼标算然后写挂了,对此表示很遗憾,但是此题确实是比较容易写挂的... 注:中国剩余定理 解模线性方程组的时候 有两种情况 1:一种是模数是两辆互质的,这样的题可以用LRJ白书上的模板,俗称CRT1 2:模数存在不互质的,这样的需要用合并方程的做法,俗称C…

链接:pid=1573">http://acm.hdu.edu.cn/showproblem.php? pid=1573 题意:求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], -, X mod a[i] = b[i], - (0 < a[i] <= 10). 思路:中国剩余定理的模板题(全部除数相互不互质版),假设找不到这种数或者最小的X大于N.输出零. 资料:http:/…

Biorhythms Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2481    Accepted Submission(s): 1091 Problem Description Some people believe that there are three cycles in a person's life that start…

Circle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description     Satiya August is in charge of souls. He finds n souls,and lets them become a circle.He ordered them to play Joseph Games.The souls will…

Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1943    Accepted Submission(s): 693 Problem Description One day I was shopping in the supermarket. There was a cashier counting coins s…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2980    Accepted Submission(s): 942 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

题意: 给定方程 res % 14 = 5 res % 57 = 56 求res 中国剩余定理裸题 #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<math.h> #include<set> #include<queue> #include<vector> using namespace s…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5446 Unknown Treasure 问题描述 On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere…

从6点看到10点,硬是没算出来,早知道玩游戏去了,艹,明天继续看 不爽,起来再看,终于算是弄懂了,以后超过一个小时的题不会再看了,不是题目看不懂,是水平不够 #include<cstdio> using namespace std; __int64 result,d; int flag; __int64 gcd(__int64 a,__int64 b,__int64 &x,__int64 &y) { __int64 t,ret; if(!b) { x = ; y = ; ret…

Unknown Treasure Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2209    Accepted Submission(s): 821 Problem Description On the way to the next secret treasure hiding place, the mathematician…

Unknown Treasure Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician…

题目链接:Lucky7 题意:求在l和r范围内,满足能被7整除,而且不满足任意一组,x mod p[i] = a[i]的数的个数. 思路:容斥定理+中国剩余定理+快速乘法. (奇+ 偶-) #include <stdio.h> #include <string.h> #include <iostream> using namespace std; #define LL long long #define FOR(i, n) for (int i=0; i<n; +…

题意: 给一个计算器,有一系列计算步骤,只有加,乘,幂三种运算. 有一种查询操作:查询初始值为\(x\)的时候,最终运算结果模\(29393\)的值. 有一种修改操作:可以修改第\(p\)个运算的运算符和运算数. 分析: 分解一下,\(29393=7 \times 13 \times 17 \times 19\). 所以我们可以维护\(4\)棵线段树,区间维护的信息就是初始值为\(x\)经过这段区间最终得到的值. 然后就用中国剩余定理整合一下. #include <cstdio> #inclu…

题意: 给定n,AA 以下n个数m1,m2···mn 则有n条方程 res % m1 = m1-AA res % m2 = m2-AA 问res的最小值 直接上剩余定理,嘿嘿 #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<math.h> #include<set> #include<queue> #i…

Hello Kiki Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 247 Accepted Submission(s): 107   Problem Description One day I was shopping in the supermarket. There was a cashier counting coins serio…

HDU 3579 Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3107    Accepted Submission(s): 1157 Problem Description One day I was shopping in the supermarket. There was a cashier counti…

题意:给你x%ci=bi(x未知),是否能确定x%k的值(k已知) ——数学相关知识: 首先:我们知道一些事情,对于k,假设有ci%k==0,那么一定能确定x%k的值,比如k=5和ci=20,知道x%20=y,那么ans=x%k=y%5; 介绍(互质版)中国剩余定理,假设已知m1,m2,mn,两两互质,且又知道x%m1,x%m2..x%mn分别等于多少 设M=m1*m2*m3..mn,那么x在模M的剩余系下只有唯一解(也就是知道了上面的模线性方程组,就可以求出x%M等于多少) ——此题解法 针对…

[欧拉函数] 在数论,对正整数n,欧拉函数是少于或等于n的数中与n互质的数的数目.此函数以其首名研究者欧拉命名,它又称为Euler’s totient function.φ函数.欧拉商数等. 例如φ(8)=4,因为1,3,5,7均和8互质. 从欧拉函数引伸出来在环论方面的事实和拉格朗日定理构成了欧拉定理的证明. [证明]: 设A, B, C是跟m, n, mn互质的数的集,据中国剩余定理,A*B和C可建立一一对应的关系.因此φ(n)的值使用算术基本定理便知, 若 n= ∏p^(α(下标p))p|…

http://poj.org/problem?id=2891 Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 11970   Accepted: 3788 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express no…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 103539   Accepted: 32012 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

http://acm.hdu.edu.cn/showproblem.php?pid=3579 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4208    Accepted Submission(s): 1617 Problem Description One day I was shopping in the supermarket.…