LightOJ 1369 - Answering Queries （预处理）

1369 - Answering Queries

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 ofA[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 between0 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

题意：不解释了，应该都能看懂

思路：预处理求出没有变化之前的ans值，然后当改变的时候：用d来表示改变量，数会有减别的数和被别的数减的情况，当减别的数时，ans变化了d*c,当被减时，ans变化了d*（n-c-1）  （ps：因为数组下边是从0开始的，所以-1），因为前面已经预处理了初始和，所以后面之需要判断ans的总变化就行了，最后改一下被修改的那个值。

ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 444440#define MOD 1000000007  #define LL long long  using namespace std;LL a[MAXN];int main(){LL t,i,n,m;int cas=0;scanf("%lld",&t);while(t--){LL sum=0;scanf("%lld%lld",&n,&m);for(i=0;i<n;i++){scanf("%lld",&a[i]);sum+=a[i];    }LL ans=0;for(i=0;i<n;i++){ans+=((n-i-1)*a[i]-(sum-a[i]));sum-=a[i];}//printf("%lld\n",ans);printf("Case %d:\n",++cas);while(m--){LL ch;scanf("%d",&ch);if(ch==1){printf("%lld\n",ans);}else if(ch==0){LL c,b;scanf("%lld%lld",&c,&b);LL d=a[c]-b;a[c]=b;ans+=(d*c-d*(n-c-1));}}}    return 0;}