LightOJ

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

4

Source

Problem Setter: Hasnain Heickal Jami

Special Thanks: Jane Alam Jan

 

细节+ 规律,

#include <cstdio>
#define N 100000+100
typedef long long LL;
LL a[N];
int main()
{
    int t, Q=1; scanf("%d", &t);
    while(t--)
    {
        int n, q; scanf("%d%d", &n, &q);
        for(int i=0; i<n; i++)
            scanf("%lld", &a[i]);
        LL getSum=0;
        //for(int i=0; i<n; i++)
        //    getSum+=(n-1-i*2)*a[i];
        LL A=n-1, b=0;
        for(int i=0; i<n; i++)
        {
            if(A <0|| b> n)
                continue;
            getSum +=A*a[i]-b*a[i];
            A--; b++; 
        }
        int firNum;
        printf("Case %d:
", Q++);
        while(q--)
        {
            scanf("%d", &firNum);
            if(firNum==1)
                printf("%lld
", getSum);
            else
            {
                int posi; LL modify;
                scanf("%d%lld", &posi, &modify);
                LL wc=modify-a[posi];
                getSum=getSum+(n-1-2*posi)*wc;
                a[posi]=modify;  //更新必须要有 !!!!! ;
            }
        }
    }
    return 0;
}