sum of power

sum of power
Time Limit: 1000MS Memory Limit: 65536KB
Submit Statistic
Problem Description

Calculate mod (1000000000+7) for given n,m.
Input

Input contains two integers n,m(1≤n≤1000,0≤m≤10).
Output

Output the answer in a single line.
Example Input

10 0
Example Output

10
Hint

Author

“浪潮杯”山东省第八届ACM大学生程序设计竞赛(感谢青岛科技大学)

注意数据范围~每一步都要取模!

#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
int main()
{
    int n, m;
    long long int sum, ans;
    while(cin>>n>>m)
    {
        ans = 0;
        for(int i=1;i<=n;i++)
        {
            sum = 1;
            for(int j=0;j<m;j++)
            {
                sum = (sum * i) % mod;
            }
            ans = (sum + ans) % mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}
Kars is tired and resentful of the narrow mindset of his village since they are content with staying where they are and are not trying to become the perfect life form. Being a top-notch inventor, Kars wishes to enhance his body and become the perfect life form. Unfortunately, n of the villagers have become suspicious of his ideas. The i -th villager has a suspicion of ai on him. Individually each villager is scared of Kars, so they form into groups to be more powerful. The power of the group of villagers from l to r be defined as f(l,r) where f(l,r)=|al−al+1|+|al+1−al+2|+…+|ar−1−ar|. Here |x−y| is the absolute value of x−y . A group with only one villager has a power of 0 . Kars wants to break the villagers into exactly k contiguous subgroups so that the sum of their power is minimized. Formally, he must find k−1 positive integers 1≤r1<r2<…<rk−1<n such that f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) is minimised. Help Kars in finding the minimum value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Input The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of test cases follows. The first line of each test case contains two integers n,k (1≤k≤n≤100) — the number of villagers and the number of groups they must be split into. The second line of each test case contains n integers a1,a2,…,an (1≤ai≤500) — the suspicion of each of the villagers. Output For each test case, output a single integer — the minimum possible value of sum of power of all the groups i. e. the minimum possible value of f(1,r1)+f(r1+1,r2)+…+f(rk−1+1,n) . Example inputCopy 3 4 2 1 3 5 2 6 3 1 9 12 4 7 2 12 8 1 9 8 2 3 3 1 8 7 7 9 2 outputCopy 4 11 2 Note In the first test case, we will group the villagers with suspicion (1,3,5,2) into (1,3,5) and (2) . So, f(1,3)+f(4,4)=(|1−3|+|3−5|)+0=4+0=4 . In the second test case, we will group the villagers with suspicion (1,9,12,4,7,2) into (1),(9,12),(4,7,2) . So, f(1,1)+f(2,3)+f(4,6)=0+3+8=11 .
07-07
标题基于SpringBoot+Vue的社区便民服务平台研究AI更换标题第1章引言介绍社区便民服务平台的研究背景、意义,以及基于SpringBoot+Vue技术的研究现状和创新点。1.1研究背景与意义分析社区便民服务的重要性,以及SpringBoot+Vue技术在平台建设中的优势。1.2国内外研究现状概述国内外在社区便民服务平台方面的发展现状。1.3研究方法与创新点阐述本文采用的研究方法和在SpringBoot+Vue技术应用上的创新之处。第2章相关理论介绍SpringBoot和Vue的相关理论基础,以及它们在社区便民服务平台中的应用。2.1SpringBoot技术概述解释SpringBoot的基本概念、特点及其在便民服务平台中的应用价值。2.2Vue技术概述阐述Vue的核心思想、技术特性及其在前端界面开发中的优势。2.3SpringBoot与Vue的整合应用探讨SpringBoot与Vue如何有效整合,以提升社区便民服务平台的性能。第3章平台需求分析与设计分析社区便民服务平台的需求,并基于SpringBoot+Vue技术进行平台设计。3.1需求分析明确平台需满足的功能需求和性能需求。3.2架构设计设计平台的整体架构,包括前后端分离、模块化设计等思想。3.3数据库设计根据平台需求设计合理的数据库结构,包括数据表、字段等。第4章平台实现与关键技术详细阐述基于SpringBoot+Vue的社区便民服务平台的实现过程及关键技术。4.1后端服务实现使用SpringBoot实现后端服务,包括用户管理、服务管理等核心功能。4.2前端界面实现采用Vue技术实现前端界面,提供友好的用户交互体验。4.3前后端交互技术探讨前后端数据交互的方式,如RESTful API、WebSocket等。第5章平台测试与优化对实现的社区便民服务平台进行全面测试,并针对问题进行优化。5.1测试环境与工具介绍测试
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值