the Sum of Cube Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 405 Accepted Submission(s): 224 Problem Description A range is given, the begin and the end are both integers. You should sum…
水 #include <stdio.h> #include <stdlib.h> #include<math.h> #include<iostream> #define LL long long using namespace std; int main() { int t; int a,b; int cas; LL sum; while(~scanf("%d",&t)) { ;i<=t;i++) { sum=; scanf…
HDU5053the Sum of Cube(水题) 题目链接 题目大意:给你L到N的范围,要求你求这个范围内的全部整数的立方和. 解题思路:注意不要用int的数相乘赋值给longlong的数,会溢出. 代码: #include <cstdio> #include <cstring> const int N = 10005; typedef long long ll; ll t[N]; void init () { for (ll i = 1; i <= N - 5; i++…
Description A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range. Input The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve. E…
Add More Zero Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 398 Accepted Submission(s): 283 Sample Input 1 64 Sample Output Case #1: 0 Case #2: 19 Source 2017 Multi-University Trainin…
题目: 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. 思路: 刚开始做写了一个尺取法,脑抽的一批.(这几天心情真的是颓的很,连带脑瓜也不好用了,抓紧调整~~~~) 其实可以根据等差数列求和公式求出这个子序列的长度的一个大致的范围为k=sqrt(2*m),然后根据m和k求出首项a1,然后将a1…
the Sum of Cube Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 162 Accepted Submission(s): 101 Problem Description A range is given, the begin and the end are both integers. You should sum t…
A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range. InputThe first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve. Each case of inp…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=4825 题面: Xor Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 6430 Accepted Submission(s): 2783 Problem Description Zeus 和 Prometheus 做了一个游戏,…
the Sum of Cube Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 1 Accepted Submission(s) : 1 Problem Description A range is given, the begin and the end are both integers. You should sum the cu…
题目链接:hdu 3415 Max Sum of Max-K-sub-sequence 题意: 给你一串形成环的数,让你找一段长度不大于k的子段使得和最大. 题解: 我们先把头和尾拼起来,令前i个数的和为sum[i]. 然后问题变成了求一个max{sum[i]-sum[j]}(i-k<j<i) 意思就是对于每一个sum[i],我们只需要找一个满足条件的最小的sum[j],然后我们就可以用一个单调队列来维护. #include<bits/stdc++.h> #define F(i,a…
HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…
