HDOJ 1097 A hard puzzle】的更多相关文章

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin. this puzzle describes that: gave a and b,how to know…
Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…
题目和1061非常相似,几乎可以复用. #include <stdio.h> ][]; int main() { int a, b; int i, j; ; i<; ++i) { buf[i][] = ; buf[i][] = i; ; j<; ++j) { buf[i][j] = buf[i][j-]*i%; ]) break; buf[i][]++; } } while (scanf("%d %d", &a, &b) != EOF) { i…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1097 分析:简单题,快速幂取模, 由于只要求输出最后一位,所以开始就可以直接mod10. /*A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33036 Accepted Submission(s): 11821 Pr…
Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…
trie树.以puzzle做trie树内存不够,从puzzle中直接找串应该会TLE.其实可以将查询组成trie树,离线做.扫描puzzle时注意仅三个方向即可. /* 1857 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector>…
Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer…
直接构造矩阵,最上面一行加一排1.高速幂计算矩阵的m次方,统计第一行的和 CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 133    Accepted Submission(s): 63 Problem Description CRB is now playing Jigsaw Puzzle. There…
dp [ x ] [ y ] [ z ] 表示二进制y所表示的组合对应的之和mod x余数为z的最小数... 如可用的数字为 1 2 3 4...那么 dp [ 7 ] [ 15 ] [ 2 ] = 1234 .... 输入一个数列后..将dp的表做出来..然后O(1)的输出...题目要求是( T + X ) % K =0 可以转化为 T % K = ( K - ( X % K ) ) % K Program: #include<iostream> #include<stdio.h>…
one recursive approach to solve hdu 1016, list all permutations, solve N-Queens puzzle. reference: the video of stanford cs106b lecture 10 by Julie Zelenski https://www.youtube.com/watch?v=NdF1QDTRkck // hdu 1016, 795MS #include <cstdio> #include &l…
Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…
A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51690 Accepted Submission(s): 18916 Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,ho…
数学归纳法,得证只需求得使18+ka被64整除的a.且a不超过65. #include <stdio.h> int main() { int i, j, k; while (scanf("%d", &k) != EOF) { j = ; ; i<; i++) { +k*i) % == ) { j = ; break; } } if (j) printf("%d\n", i); else printf("no\n"); }…
HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…
这道题就是HDOJ的1061的变形: 1061 :求n的n次方的个位数 http://www.cnblogs.com/xiezie/p/5596779.html 1097 :求n的m次方的个位数 因此,就不在这里赘述了 以下是JAVA实现: import java.io.BufferedInputStream; import java.util.ArrayList; import java.util.Scanner; public class Main { public static void…
HDOJ 题目分类 /* * 一:简单题 */ 1000:    入门用:1001:    用高斯求和公式要防溢出1004:1012:1013:    对9取余好了1017:1021:1027:    用STL中的next_permutation()1029:1032:1037:1039:1040:1056:1064:1065:1076:    闰年 1084:1085:1089,1090,1091,1092,1093,1094, 1095, 1096:全是A+B1108:1157:1196:1…
Play on Words Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5622    Accepted Submission(s): 1850 Problem Description Some of the secret doors contain a very interesting word puzzle. The team…
Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9934    Accepted Submission(s): 6959 Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no…
Ignatius's puzzle Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5x13+13*x5+ka*x,input a nonegative integer k(k<10000),to find the minimal none…
FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…
Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…
Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…
Puzzle 面向服务/切面AOP开发框架 For .Net AOP主要实现的目的是针对业务处理过程中的切面进行提取,它所面对的是处理过程中的某个步骤或阶段,以获得逻辑过程中各部分之间低耦合性的隔离效果. 日常的产品开发中最常见的就是数据保存的功能.举例来说,现在有个用户信息数据保存的功能,我们希望在数据保存前对数据进行校验,其中现在能想到的校验就包含数据完整性校验和相同数据是否存在校验.按照传统的OOP(面向对象程序设计),我们需要定义一个IUserDataSaveService(用户数据保存…
题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5456 Description As an exciting puzzle game for kids and girlfriends, the Matches Puzzle Game asks the player to find the number of possible equations A−B=C with exactly n (5≤n≤500) matches (or stic…
Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…
卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…
Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…
Katu Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6714   Accepted: 2472 Description Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integ…
Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8737   Accepted: 5468 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…
前言 不说话,先猛戳 Ranklist 看我排名. 这是用 node 自动刷题大概半天的 "战绩",本文就来为大家简单讲解下如何用 node 做一个 "自动AC机". 过程 先来扯扯 oj(online judge).计算机学院的同学应该对 ACM 都不会陌生,ACM 竞赛是拼算法以及数据结构的比赛,而 oj 正是练习 ACM 的 "场地".国内比较有名的 oj 有 poj.zoj 以及 hdoj 等等,这里我选了 hdoj (完全是因为本地上…