链接:http://poj.org/problem?id=2769 题意:寻找数m,是的对于n个数的余数不同 思路:暴力,优化:同余部分不用测试 代码: #include <iostream> #include <string.h> using namespace std; ],vis[]; int main() { ios::sync_with_stdio(false); //freopen("in.txt","r",stdin); //f…

Description T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 106-1. T. Chur finds this range of SINs too larg…

T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 10 6-1. T. Chur finds this range of SINs too large for ident…

Description T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 10 6-1. T. Chur finds this range of SINs too lar…

  http://poj.org/problem?id=1651Multiplication Puzzle   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6188   Accepted: 3777 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer.…

http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23339   Accepted: 8284 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree…

http://poj.org/problem?id=3278 题目大意: m,n两个数m可+1, -1, *2变成n,需要经过几步 #include<stdio.h> #include<stdlib.h> #include<math.h> #include<string.h> #include<queue> #define max(a, b)(a > b ? a : b) #define N 100010 using namespace s…

POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 3669 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Sciss…

http://poj.org/problem?id=2796 题意:求n个数的和乘以这n个数中的最小值的积最大的数,以及其范围. 思路:求每一个数两边的比其大的数的和,再乘以这个数.还有一个范围,用单调栈找以及记录. 这个题和2559差不多,就是多了一个对数字的求和. #include <stdio.h> #include <iostream> #include <stack> #define x 100010 using namespace std; stack<…

http://poj.org/problem?id=3056 The Bavarian Beer Party Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 995   Accepted: 359 Description The professors of the Bayerische Mathematiker Verein have their annual party in the local Biergarten.…

poj 2769 感觉♂良好 (单调栈) 比尔正在研发一种关于人类情感的新数学理论.他最近致力于研究一个日子的好坏,如何影响人们对某个时期的回忆. 比尔为人的一天赋予了一个正整数值. 比尔称这个值为当天的情感价值.情感价值越大,日子过的越好.比尔认为,某个时期人的情感,与某一时期的情感价值总和,乘以这一阶段的最小情感价值成正比.这个想法表示,一段时期的好心情可以被糟糕的一天破坏. 现在比尔正在审视自己的生活,并希望找到最有价值的人生阶段.请你帮帮他.由于他过度♂劳累,他的生命天数n<=10000…

题目链接:http://poj.org/problem?id=1995 解题思路:用整数快速幂算法算出每一个 Ai^Bi,然后依次相加取模即可. #include<stdio.h> long long quick_mod(long long a,long long b,long long c) { long long ans=1; while(b) { if(b&1) { ans=ans*a%c; } b>>=1; a=a*a%c; } return ans; } int…

题目链接 http://poj.org/problem?id=1430 题解 qaq写了道水题-- 在模\(2\)意义下重写一下第二类Stirling数的递推式: \[S(n,m)=S(n-1,m-1)+(S(n-1,m)\ \text{and}\ m)\] 令\(S'(n,m)=S(n+m,m)\), 那么递推式变成了\(S'(n,m)=S'(n,m-1)+(S'(n-1,m)\ \text{and}\ m)\) 也就相当于从\((0,0)\)走到\((n,m)\)的NE Lattice Pa…

嗯... 题目链接:http://poj.org/problem?id=1995 快速幂模板... AC代码: #include<cstdio> #include<iostream> using namespace std; int main(){ ; scanf("%lld", &N); while(N--){ scanf("%lld%lld", &M, &n); sum = ; ; i <= n; i++){…

题目链接 http://codeforces.com/gym/101102/problem/J Description standard input/output You are given an array A of integers of size N, and Q queries. For each query, you will be given a set of distinct integers S and two integers L and R that represent a…

Binary Stirling Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1761   Accepted: 671 Description The Stirling number of the second kind S(n, m) stands for the number of ways to partition a set of n things into m nonempty subsets.…

ID Codes Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6281 Accepted: 3769 Description It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby…

                               Smith Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14521   Accepted: 4906 Description While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed tha…

Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6347   Accepted: 3740 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

memset可以对高位数组进行初始化,非常方便.需要注意的是memset的头文件是string.h和memory.h . 下面来谈memset的4个使用技巧: (注:一下dp高维数组都是全局变量,局部变量请自行修改sizeof语句) 1. 用memset赋 0 memset(dp,,sizeof(dp)); 2. 用memset赋 -1 memset(dp,-,sizeof(dp)); 在计算机中,数据用补码保存.-1的补码(32位)是0xFFFFFFFF,(8位,一个字节)是0xFF,mems…

Pseudoprime numbers Descriptions 费马定理指出,对于任意的素数 p 和任意的整数 a > 1,满足 ap = a (mod p) .也就是说,a的 p 次幂除以 p 的余数等于 a .p 的某些 (但不是很多) 非素数的值,被称之为以 a 为底的伪素数,对于某个 a 具有该特性.并且,某些 Carmichael 数,对于全部的 a 来说,是以 a为底的伪素数. 给定 2 < p ≤ 1000000000 且 1 < a < p ,判断 p 是否为以 …

Street Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2753   Accepted: 1530 Description A computer programmer lives in a street with houses numbered consecutively (from 1) down one side of the street. Every evening she walks her…

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 76935   Accepted: 24323 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,00…

题意:给两个整数,求这两个数的反向数的和的反向数,和的末尾若为0,反向后则舍去即可.即若1200,反向数为21.题目给出的数据的末尾不会出现0,但是他们的和的末尾可能会出现0. #include <iostream> #include <string.h> #include <stdio.h> #include <string> #include <string.h> using namespace std; int n1,n2;//n1表示a的…

题意:在墙上贴海报,海报可以互相覆盖,问最后可以看见几张海报思路:直接搞超时+超内存,需要离散化.离散化简单的来说就是只取我们需要的值来 用,比如说区间[1000,2000],[1990,2012] 我们用不到[-∞,999][1001,1989][1991,1999][2001,2011][2013,+∞]这些值,所以我只需要 1000,1990,2000,2012就够了,将其分别映射到0,1,2,3,在于复杂度就大大的降下来了所以离散化要保存所有需要用到的值,排序后,分别映射到1~n,这样复…

Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5477   Accepted: 3173 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

#include <iostream>// poj 1258 10.1.5.253 1505 using namespace std; #define N 105 // 顶点的最大个数 (多写 int a[N][N],low[N],n,ans; int min(int x,int y) { return x<y?x:y; } void prim(int u0) { int i,j,m,k; ans=0; // for (i=1;i<n;i++) low[i]=a[u0][i]; /…

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked…

Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 9512 Accepted: 5783 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others…

题目链接 #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<string> using namespace std; ; typedef long long int LL; LL sum[N << ]; LL add[N << ]; struct node { int…