Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24116    Accepted Submission(s): 10232 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],…

Number Sequence Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Problem Description There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules: ● a i ∈ [0,n] ● a i ≠ a j ( i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1005 Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 85249    Accepted Submission(s): 20209 Problem Description A number sequence…

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 89984    Accepted Submission(s): 21437 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A…

Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multiple test cases. Each test case…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5064 题目意思:给出n个数:a1, a2, ..., an,然后需要从中找出一个最长的序列 b1, b2, ...,  bt.需要满足两个条件(1)b1≤b2≤…≤bt   (2)b2−b1≤b3−b2≤⋯≤bt−bt−1.求出最大的 t 为多少. 遗留大半年的题目呀呀呀呀~~~~!!!首先非常感谢乌冬子大半年前的指点迷津,呕心沥血想了大半天举出个反例,以便指出我算法的错误.为了纪念这个伟大的人物(…

http://acm.hdu.edu.cn/showproblem.php?pid=1711 这道题就是一个KMP模板. #include<iostream> #include<cstring> using namespace std; +; int n,m; int next[maxn]; int a[maxn], b[maxn]; void get_next() { , j = ; ::next[] = -; while (j < m) { || b[i] == b[j]…

#include <stdio.h> #include <string.h> int main() { int a,b,n; int i; ]={}; f[]=; f[]=; while(scanf("%d %d %d",&a,&b,&n)!=EOF){ ){ break; } ;i<;i++){ f[i]=(a*f[i-]+b*f[i-])%; } ){ printf(]); } else{ printf()%+]); } } ;…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5014 解题报告:西安网赛的题,当时想到一半,只想到从大的开始匹配,做异或运算得到对应的b[i],但是少了一个操作,ans[i] = temp,却没有想到ans[temp] = i;所以就一直卡在这里了,因为我不确定这样是不是一一对应的,好吧,也没有想到这里,就差这么点了. #include<cstdio> #include<cstring> #include<iostream&g…

链接:传送门 题意:略 思路:f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7 -> f(n) = (A * f(n-1)%7 + B * f(n-1)%7)%7 检查输出可以发现规律,每48个数一循环,因此只需要打出前50个数的fib表然后对n%7即可. /************************************************************************* > File Name: 1.cpp > Aut…

题意,f(1)=1,f(2)=1,f(n)=a*f(n-1)+b*f(n-2),求f(n)%7 这个题可能数据不够严谨,所以有些错误的做法也可以通过,比如7 7 50,应该输出0而不是1 解:找到关键字%7,那么能说明每一步都%7,最终答案也就是%7的,如果每一步都%7,那么对于任意f(n)来说,f(n-1)有7种情况,f(n-2)有7种情况,一共最多只有49种情况,说明这里有循环节,而且循环节的长度不超过49.需要注意的是,这里只能证明循环节的长度不超过49,并不能证明f(1)和f(2)就在循…

题意:给出一串数列,这串数列的gcd为1,要求取出一个数使取出后的数列gcd最大. 题解:可以通过对数列进行预处理,求出从下标为1开始的数对于前面的数的gcd(数组从下标0开始),称为前缀gcd,再以类似的方式求出后缀gcd,然后从第一个数开始枚举取出后的gcd(这个数的前缀gcd与后缀gcd的gcd).找出最大的gcd数即可. 代码如下: #include<cstdio> #include<cstring> #include<algorithm> using name…

/* 题意:a, b两个序列,规定由[0, n]区间的数! 求 a[i] ^ b[i] 的和最大! 思路:如果数字 n的二进制有x位, 那么一定存在一个数字m,使得n^m的所有二进制位 都是1,也就是由x位1!这样下去的到的值就是最大值! */ #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define N 100005 using namespace s…

题意:给定一个字符串,I表示本字符要比前一个字符大,D表示本字符要不前一个字符小,?可大可小,问1~n的所有排列中,有多少满足条件 /* dp方程的设定比较显然,dp[i][j]表示选了i个元素,最后一个是j的方案数. 但是在状态转移的时候,我们不得不考虑前面选了什么,也就是状态的设定是有后效性的, 所以考虑给状态再添一层含义:必须选前i个元素. 那么这样岂不是每次只能选i吗?那么第二维岂不是没有用了? 所以我们考虑用j把i替换出来,那么在状态转移的时候就需要考虑放入i时,怎么替换能使原来的大小…

题意: 有一串数字串,其规律为 1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011 123456789101112······k 输入位置n,计算这一串数字第n位是什么数字,注意是数字,不是数! 例如12345678910的第10位是1,而不是10,第11位是0,也不是10.总之多位的数在序列中要被拆分为几位数字,一个数字对应一位. //运用了一个公式:求一个数的位数时,可以用 (int)…

HDU 1711 Number Sequence(数列) Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) [Description] [题目描述] Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N…

HDU 1005 Number Sequence(数列) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, an…

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HDU 1560 DNA sequence(DNA序列) Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)   Problem Description - 题目描述 The twenty-first century is a biology-technology developing century. We know that a gene is made of DNA. Th…

1085 Perfect Sequence (25 分) Given a sequence of positive integers and another positive integer p. The sequence is said to be a perfect sequence if M≤m×p where M and m are the maximum and minimum numbers in the sequence, respectively. Now given a seq…

1140 Look-and-say Sequence(20 分)Look-and-say sequence is a sequence of integers as the following: D, D1, D111, D113, D11231, D112213111, ... where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, th…

//1088(参考博客:http://blog.csdn.net/libin56842/article/details/8950688)//1.编写一个浏览器输入输出(hdu acm1088)://思路:对已经输入的字符串进行处理,遇到<br><hr>分别进行处理.遇到多于80个字符(统计该行的长度)或者<br>或者结束,则换行:遇到<hr>,输出80个'-'.#include<stdio.h>#include<string.h>#d…

一.使用场景 当需要比较范围时 如: 这种情况,如果要写三个表达式会很长,这时候就可以用这个工具类进行比较 number:用户输入(长,宽,高) minRange: 0.0 maxRange:33 二.判断number存在(minRange ~ maxRange)范围中 /** * 判断number存在(minRange ~ maxRange)范围中 * * @param number * @param minRange * @param maxRange * @return */ public…

题目代号:HDU 1711 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711 Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 28288    Accepted Submission(s): 11891 Problem Description Give…

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39408    Accepted Submission(s): 16269 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],…

题目链接 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M…

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 40388    Accepted Submission(s): 16659 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],…

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000…

HDU - 1711 A - Number Sequence   Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... ,…