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1369 - Answering Queries(规律)
阅读量:623 次
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

1

3 5

1 2 3

1

0 0 3

1

0 2 1

1

Case 1:

-4

0

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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