lightoj 1369 - Answering Queries 【思维】

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

题意：给定f(A, n)函数的程序 和 数组A[]，现在有q次操作。

操作有两种一、1 表示求f(A, n)的值；二、0 x v 表示把A[x] 修改为 v。 要求每次查询时间复杂度为O(1)。

思路：预处理初始的f(A, n) = ans，考虑每个A[i]做出的贡献。

可以求出A[i]的贡献为A[i] * (n-i-1) - sigma(A[i] + ... + A[n-1])，该操作可以在常数时间内完成。求出ans的时间复杂度为O(n)。

对于每次修改A[x] = v，考虑该元素修改对ans的影响。前面元素使它减少(v-A[x])*x，它对后面的影响使它减少(A[x]-v)*(n-x-1)。 这样 可得ans -= (v-A[x])*x + (A[x]-v)*(n-x-1)。

这样修改和查询都可以在常数时间内完成。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 100000+10
#define MAXM 50000000
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
LL A[MAXN];
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        int n, q;
        Ri(n); Ri(q);
        LL sum = 0;
        for(int i = 0; i < n; i++)
        {
            Rl(A[i]);
            sum += A[i];
        }
        sum -= A[0];
        LL ans = 0;
        for(int i = 0; i < n-1; i++)
        {
            ans += A[i] * (n-i-1) - sum;
            sum -= A[i+1];
        }
        printf("Case %d:\n", kcase++);
        while(q--)
        {
            int op; Ri(op);
            if(op == 1)
                Pl(ans);
            else
            {
                int x, v;
                Ri(x); Ri(v);
                ans -= (v-A[x]) * x + (A[x]-v) * (n-x-1);
                A[x] = v;
            }
        }
    }
    return 0;
}