1.链接地址: http://bailian.openjudge.cn/practice/1844 http://poj.org/problem?id=1844 2.题目: Sum Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10031   Accepted: 6564 Description Consider the natural numbers from 1 to N. By associating to eac…

链接: http://poj.org/problem?id=1844 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29256#problem/D Sum Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 9795   Accepted: 6406 Description Consider the natural numbers from 1 to N. By a…

题意:给一个整数n,求当n由1到k的连续整数加或减组成时的最小的k. 解法:一开始觉得dp……后来觉得复杂度太大了……GG……百度了一下是个数学题orz. 如果n全部由加法组成,那么k可以组成k(k+1)/2,设减掉的部分为s,则有k(k+1)/2-2s=n,所以n和k(k+1)/2mod2同余. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #in…

POJ 2739 Sum of Consecutive Prime Numbers Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu   Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representati…

POJ.2739 Sum of Consecutive Prime Numbers(水) 代码总览 #include <cstdio> #include <cstring> #include <vector> #include <cmath> #define nmax 10005 using namespace std; int prime[nmax],n; vector<int> vprime; void init() { memset(pri…

POJ 2739 Sum of Consecutive Prime Numbers(素数) http://poj.org/problem? id=2739 题意: 给你一个10000以内的自然数X.然后问你这个数x有多少种方式能由连续的素数相加得来? 分析: 首先用素数筛选法把10000以内的素数都找出来按从小到大保存到prime数组中. 然后找到数X在prime中的上界, 假设存在连续的素数之和==X, 那么一定是从一个比X小的素数開始求和(不会超过X的上界),直到和sum的值>=X为止. 所…

Sum Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10494   Accepted: 6895 Description Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

题目链接:http://poj.org/problem?id=2739 预处理出所有10001以内的素数,按照递增顺序存入数组prime[1...total].然后依次处理每个测试数据.采用双重循环计算n的表示数: 外循环i :  for (i = 0; x >= prime[i]; i++) 的循环结构枚举所有可能的最小素数prime[i]: 内循环:   while (ans < x && j < total)   ans += prime[j++];    计算连续…

题目连接 http://poj.org/problem?id=1564 Sum It Up Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then…

http://poj.org/problem?id=1564 该题运用DFS但是要注意去重,不能输出重复的答案 两种去重方式代码中有标出 第一种if(a[i]!=a[i-1])意思是如果这个数a[i]和上一个数相同,那么记录数组的同一个位置就没有必要再放入这个数.例如:4 3 3 2构成和是7,b数组的第二个位置放了3,则后面的那个3就没有必要再放入记录数组的第二个位置了.(可能会放到后面的位置)... #include<stdio.h> #include<string.h> #i…

Sum It Up Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…

http://poj.org/problem?id=1775 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1334 题目大意: 给一个数n看看n是否能够拆成几个阶乘的和 如9=1!+2!+3! 方法一: 最初想法是直接打出0~10的阶乘,10的阶乘已经大于n的范围 然后DFS,也过了.不过时间好惨... 注意n=0输出NO和0!=1 #include<cstdio> #include<cstring> con…

题目链接: http://poj.org/problem?id=1707 Language: Default Sum of powers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 735   Accepted: 354 Description A young schoolboy would like to calculate the sum   for some fixed natural k and differe…

题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS     Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive intege…

题目链接:http://poj.org/problem?id=2739 #include <cstdio> #include <cstring> using namespace std; int method[10001][1300]; int dp[10001]; bool isntprime[10001]; int heap[1300],cnt; void calprime(){ method[2][0]=1; dp[2]++; heap[cnt++]=2; for(int i…

题目poj 题目zoj //我感觉是题目表述不确切,比如他没规定xi能不能重复,比如都用1,那么除了0,都是YES了 //算了,这种题目,百度来的过程,多看看记住就好 //题目意思:判断一个非负整数n能否表示成几个数的阶乘之和 //这里有一个重要结论:n!>(0!+1!+……+(n-1)!), //证明很容易,当i<=n-1时,i!<=(n-1)!,故(0!+1!+……+(n-1)!)<=n*(n-1)!=n!. // 由于题目规定n<=1000000,而10!=362880…

题目链接:http://poj.org/problem?id=1707 题意:给出n 在M为正整数且尽量小的前提下,使得n的系数均为整数. 思路: i64 Gcd(i64 x,i64 y) { if(y==0) return x; return Gcd(y,x%y); } i64 Lcm(i64 x,i64 y) { x=x/Gcd(x,y)*y; if(x<0) x=-x; return x; } struct fraction { i64 a,b; fraction() {} fractio…

Sum It Up Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6682   Accepted: 3475 Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n…

Sum It Up Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if…

Maximum sum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33918   Accepted: 10504 Description Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: Your task is to calculate d(A). Input The input consists o…

 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 5…

                                                                                                               Sum It Up Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5820   Accepted: 2970 Description Given a specified total t and a…

Time Limit: 1000MS   Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two r…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19697   Accepted: 10800 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

                                                                                                         Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24498   Accepted: 13326 Description Some positive i…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19895   Accepted: 10906 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

题意: 给出一个数T,再给出n个数.若n个数中有几个数(可以是一个)的和是T,就输出相加的式子.不过不能输出相同的式子. 分析: 运用的是回溯法.比较特殊的一点就是不能输出相同的式子.这个可以通过map来实现:map<string,int>把字符串(可以是C语言的字符串)和整数联系起来了.我们可以把相加起来的几个数变成一个字符串(2+1+1,变成“211”),如果它出现过,就标记为1,初始值是0.出现过的就不再输出了. 剪枝: 1.所有的数加起来的和小于T,直接输出NONE. 2.搜索过程中,…

解题思路:给定一个数,判定它由几个连续的素数构成,输出这样的种数 用的筛法素数打表 Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20020   Accepted: 10966 Description Some positive integers can be represented by a sum of one or more consecutive…

Sum It Up Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7191   Accepted: 3745 Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n…