1369 - Answering Queries<数>

最新推荐文章于 2021-06-22 16:57:05 发布

原创最新推荐文章于 2021-06-22 16:57:05 发布 · 343 阅读

0 ·

CC 4.0 BY-SA版权

数学同时被 3 个专栏收录

59 篇文章

订阅专栏

模拟

13 篇文章

订阅专栏

light oj

12 篇文章

订阅专栏

本文深入探讨了大数据开发领域的关键技术和应用，包括Hadoop、Spark、Flink等主流框架，以及数据处理、存储、分析等核心环节，旨在为读者提供全面的大数据开发知识体系。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

E - Answering Queries
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit

Status
Description
The problem you need to solve here is pretty simple. You are give a function f(A, n), where A is an array of integers and n is the number of elements in the array. f(A, n) is defined as follows:

long long f( int A[], int n ) { // n = size of A
long long sum = 0;
for( int i = 0; i < n; i++ )
for( int j = i + 1; j < n; j++ )
sum += A[i] - A[j];
return sum;
}
Given the array A and an integer n, and some queries of the form:

1) 0 x v (0 ≤ x < n, 0 ≤ v ≤ 106), meaning that you have to change the value of A[x] to v.
2) 1, meaning that you have to find f as described above.

Input
Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers: n and q (1 ≤ n, q ≤ 105). The next line contains n space separated integers between 0 and 106 denoting the array A as described above.

Each of the next q lines contains one query as described above.

Output
For each case, print the case number in a single line first. Then for each query-type "1" print one single line containing the value of f(A, n).

Sample Input
1
3 5
1 2 3
1
0 0 3
1
0 2 1
1
Sample Output
Case 1:
-4
0

血一样的教训，，，，，以后light oj 只用long long ,,,,打死也不用__int64

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
LL n,q,shu[101000];
LL s,sum;
int main()
{
	int t;
	LL a,c,ca,b,i;
	scanf("%d",&t);
 //   int t;scanf("%d",&t);int a,b,c;
	for (ca=1;ca<=t;ca++)
	{
		scanf("%lld%lld",&n,&q);
		memset(shu,0,sizeof(shu));
		s=0;sum=0;
		for (i=0;i<n;i++)
		{
			scanf("%lld",&shu[i]);
			//s+=shu[i];
			sum=sum+(n-1-i)*shu[i]-i*shu[i];
		}
		printf("Case %lld:\n",ca);
		/*for (i=0;i<n-1;i++)
		{
			s-=shu[i];
			sum=sum+(n-1-i)*shu[i]-s;
		}*/
		while (q--)
		{
			scanf("%lld",&a);
			if (a==1)
			printf("%lld\n",sum);
			else
			{
				scanf("%lld%lld",&b,&c);
				a=c;
				c=c-shu[b];
				shu[b]=a;
				sum=sum+(n-1-b)*c-b*c;
			}
		}
	}
	return 0;
}