Sum Problem Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 338086    Accepted Submission(s): 85117 Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this pro…

#include <stdio.h> int main(){ int k,sum; while(scanf("%d",&k)!=EOF){ ==){ sum=(+k)*(k/); } else{ sum=(+k)*(k/)+k/+; } printf("%d\n\n",sum); } ; }…

/* 注意可以是负整数,而且在过程中会超过int,所以要用longlong */ #include <cstdio> int main() { long long n; while (scanf("%lld",&n)!=EOF) (n<1)?printf("%d\n\n",(n+1)*(2-n)/2):printf("%d\n\n",n*(n+1)/2); return 0; }…

这一段时间一直都在刷OJ,这里建一个博客合集,用以记录和分享算法学习的进程. github传送门:https://github.com/haoyuanliu/Online_Judge/tree/master/HangDianOJ   Problem Description In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. Input The input will consist of a series…

这一套题做错了几次,按理说直接用等差数列求和公式就行了,主要是要考虑一些运算符的结核性问题: 四则运算符(+.-.*./)和求余运算符(%)结合性都是从左到右. 于是,我自己写了一个版本,主要是考虑(n+1)*n始终为偶数,这样就不用担心除以2时的取整问题: #include <stdio.h> int main(void) { int n; __int64 sum = ; while (scanf("%d", &n) != EOF) { sum = ((n + )…

The sum problem Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.   Input Input contains multiple test cases. each case contains two integers N, M( 1 <=…

The sum problem Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 31453    Accepted Submission(s): 9414 Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possib…

Sum Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=4704 Mean: 给定一个大整数N,求1到N中每个数的因式分解个数的总和. analyse: N可达10^100000,只能用数学方法来做. 首先想到的是找规律.通过枚举小数据来找规律,发现其实answer=pow(2,n-1); 分析到这问题就简单了.由于n非常大,所以这里要用到费马小定理:a^n ≡ a^(n%(m-1)) * a^(m-1)≡ a^(n%(m-…

题目链接:hdu 5106 Bits Problem 题目大意:给定n和r,要求算出[0,r)之间全部n-onebit数的和. 解题思路:数位dp,一个ct表示个数,dp表示和,然后就剩下普通的数位dp了.只是貌似正解是o(n)的算法.可是n才 1000.用o(n^2)的复杂度也是够的. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long…

HDOJ(HDU).1258 Sum It Up (DFS) [从零开始DFS(6)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DF…