Light OJ 1369 - Answering Queries 【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.

恩，题目大意就是说，定义一函数为计算 i 从1到 n-1 ，a [ i ] - a[ k ] ( i < k < n ) 之和，然后有 q 个操作，0 x v 代表将数组下标为 x 的数改为 v ，1 代表计算此时的函数值。然后直接计算肯定超时，数据太大，之和就是 a0-a1 + a0-a2 + a0-a3 + a0-a4 .......然后比较（i=1时） a1-a2 + a1-a3 + a1-a4 ...... 发现第一个式子刚好比第二个多出来 n-i 个（a0-a1)。恩，之后操作时要改变值时直接在之前求得的总和身上改，例如改变了 a7 增加了2 则从第8到n 个数被 a7减时结果都要多2，而从第0个到第6个减 a7 时结果都小2 ，就这样。

#include <iostream>#include<cstdio>#include<cstring>#define maxn 100000+10using namespace std;int a[maxn];long long f(int a[],int n){    long long s=0,tem=0;    for(int j=1;j<n;++j)        tem+=a[0]-a[j];    s=tem;    for(int i=1;i<n;++i)    {        long long t=a[i]-a[i-1];        tem=tem+t*(n-i);        s+=tem;    }    return s;}int main(){    int t,n,q,m,s,c,cnt=0;    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",++cnt);        long long sum=f(a,n);        while(q--)        {            scanf("%d",&c);            if(c==1)                printf("%lld\n",sum);            else            {                scanf("%d%d",&m,&s);                long long tem=s-a[m];                a[m]=s;                sum+=(n-2*m-1)*tem;            }        }    }    return 0;}