zoj The 12th Zhejiang Provincial Collegiate Programming Contest Beauty of Array
2024-10-12 21:28:36
http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5496
The 12th Zhejiang Provincial Collegiate Programming Contest - D
Beauty of Array
Time Limit: 2 Seconds Memory Limit: 65536 KB
Time Limit: 2 Seconds Memory Limit: 65536 KB
Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the array A.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer N (1 <= N <= 100000), which indicates the size of the array. The next line contains N positive integers separated by spaces. Every integer is no larger than 1000000.
Output
For each case, print the answer in one line.
Sample Input
3
5
1 2 3 4 5
3
2 3 3
4
2 3 3 2
Sample Output
105
21
38 分析: 水题、
AC代码:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 1000005
#define mod 786433
#define exp 1e-5
const int INF = 0x3f3f3f3f; LL dp[Len];
int hsh[Len]; int main()
{
int t,n,a;
scanf("%d",&t);
w(t--)
{
memset(hsh,,sizeof(hsh));
scanf("%d",&n);
dp[]=;
for(int i=;i<=n;i++)
{
scanf("%d",&a);
dp[i]=dp[i-]+a+(i--hsh[a])*a;
hsh[a]=i;
}
LL ans=;
for(int i=;i<=n;i++)
ans+=dp[i];
printf("%lld\n",ans);
}
}
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;
long long int dp[];
int a[];
int main() {
int t;
scanf("%d",&t);
while(t--){
int n; scanf("%d",&n);
map<int,int> num; //num记录前一个a[i]的位置 for(int i = ;i < n;i++){
scanf("%d",&a[i]);
num[a[i]] = ;
}
dp[] = ;
dp[] = a[];
num[a[]] = ; for(int i = ;i < n;i++){
/*
dp[i] 为前i个美丽的数组的和
前i个和包括前i-1个的和 dp[i-1] 还有由第i个数组成的和 dp[i] - dp[i-1] + a[i] * (i+1 - num[a[i]]);
其中的i+1 - num[a[i]] 是用于去除重复的a[i],重复的只计算一次
*/
dp[i+] = dp[i] + dp[i] - dp[i-] + a[i] * (i+ - num[a[i]]);
num[a[i]] = i+; //将新的位置记录下来
}
cout << dp[n] << endl;
}
return ;
}
另外附上本次比赛的终榜:(1-104)
最新文章
- spring mvc 4.3.2 + mybatis 3.4.1 + mysql 5.7.14 +shiro 幼儿园收费系统 之 动态组合条件查询
- flash
- git merge之squash
- Spring基础——在 Spring Config 文件中基于 XML 的 Bean 的自动装配
- 7.python模块补充
- 单源最短路径的Bellman-Ford 算法
- Linux kernel AACRAID Driver Compat IOCTL 本地安全绕过漏洞
- service redis does not support chkconfig的解决办法
- 02、Unicode 汉子转码小工具
- WinForm 控件(下)
- 2017-07-05 (whereis which find)
- svn提交代码时不要提交bulid里的内容,会报错
- C# 创建Dll文件供程序调用方法
- Failed to execute goal org.apache.maven.plugins:maven-compiler-plugin:3.1:compile (default-compile) on project ppcloud-common: Compilation failure
- deno学习四 docker 运行官方的一个http file server
- Liunx read
- 删除php数组中的元素
- java spring 等启动项目时的异常 或 程序异常的解决思路
- PHP命名空间学习笔记
- kubernetes外部访问的几种方式