1369 - Answering Queries

1369 - Answering Queries
Time Limit: 3 second(s) Memory Limit: 32 MB

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

Output for Sample Input

1

3 5

1 2 3

1

0 0 3

1

0 2 1

1

Case 1:

-4

0

4

Note

Dataset is huge, use faster I/O methods.

PROBLEM SETTER: HASNAIN HEICKAL JAMI
SPECIAL THANKS: JANE ALAM JAN
思路：推导下公式就行：sum=（n-2*i+1）*bns[i]；然后改变的时候直接改变就行，减去原来的加上现在的；
 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<stdlib.h>
 7 typedef long long LL;
 8 LL bns[200000];
 9 int main(void)
10 {
11     int k;
12     int i,j;
13     scanf("%d",&k);
14     int s;
15     int p,q;
16     LL ans=0;
17     for(s=1; s<=k; s++)
18     {
19         scanf("%d %d",&p,&q);
20         for(i=1; i<=p; i++)
21         {
22             scanf("%lld",&bns[i]);
23         }
24         ans=0;
25         for(i=1; i<=p; i++)
26         {
27             ans+=(LL)(p-2*i+1)*(LL)bns[i];
28         }
29         printf("Case %d:\n",s);
30         while(q--)
31         {
32             int ask;
33             int n,m;
34             scanf("%d",&ask);
35             if(ask==1)
36             {
37                 printf("%lld\n",ans);
38             }
39             else
40             {
41                 scanf("%d %d",&n,&m);
42 
43                 {
44                     ans-=(LL)(p-2*(n+1)+1)*bns[n+1];
45                     ans+=(LL)(p-2*(n+1)+1)*(LL)m;
46                     bns[n+1]=m;
47                 }
48             }
49         }
50     }
51     return 0;
52 }

 

posted @ 2016-05-29 11:52  sCjTyC  阅读(142)  评论(0)  编辑  收藏  举报