1369 - Answering Queries（规律）

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发布时间：2019-03-13

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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

3 5

1 2 3

0 0 3

0 2 1

Case 1:

-4

题解：我的思路本来是针对每次的修改，都在询问里面找值，不出意外肯定超时了，出来看了大神的题解，是针对每次修改再修改sum就妥了，比赛的时候就没想到。。。

代码：

1 #include
    
      2 #include
     
       3 #include
      
        4 #include
       
         5 #include
        
          6 #define mem(x,y) memset(x,y,sizeof(x)) 7 using namespace std; 8 const int INF=0x3f3f3f3f; 9 const int MAXN=1e5+100;10 typedef long long LL;11 LL a[MAXN],b[MAXN];12 int main(){13     int T,n,q,cnt=0;14     scanf("%d",&T);15     while(T--){16         scanf("%d%d",&n,&q);17         LL sum=0;18         for(int i=0;i
         
          =1;i--)sum+=a[i],b[i-1]=sum;21 //for(int i=0;i

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