lightoj 1369 - Answering Queries (思维规律)

1369 - Answering Queries

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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.

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PROBLEM SETTER: HASNAIN HEICKAL JAMI

SPECIAL THANKS: JANE ALAM JAN

1.a[0]-……

2.a[1]-……

1比2 多了（n-i）个a[0]-a[1]

类推，

这样只要一次循环就好

#include<cstdio>int a[100000+10];long long f(int a[],int n){long long s=0,tem=0;for(int i=1;i<n;i++)tem+=a[0]-a[i];s=tem;for(int i=1;i<n;i++){long long t=a[i]-a[i-1];//必须先化成long long 型 tem+=(n-i)*t;s+=tem;}return s;}int main(){int t,n,q;int k=1;int flag,x,v;scanf("%d",&t);while(t--){scanf("%d%d",&n,&q); for(int i=0;i<n;i++){scanf("%d",&a[i]);}printf("Case %d:\n",k++);long long sum=f(a,n);while(q--){scanf("%d",&flag);if(flag==1){printf("%lld\n",sum);}else{scanf("%d%d",&x,&v);long long tem=(v-a[x]);sum+=(n-2*x-1)*tem;a[x]=v;}}}return 0;}