LightOJ - 1369 - Answering Queries（规律）

链接：

https://vjudge.net/problem/LightOJ-1369

题意：

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.
   

思路：

找规律，计算每个位置的贡献。

a[i]的贡献 = (n-1-i)a[i]-ia[i];

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>

using namespace std;
typedef long long LL;
const int INF = 1e9;

const int MAXN = 1e5+10;
const int MOD = 1e9+7;

LL A[MAXN];
int n, q;

LL f(LL A[], int n)
{
    LL sum = 0;
    for (int i = 0;i < n;i++)
    {
        for (int j = i+1;j < n;j++)
            sum += A[i]-A[j];
    }
    return sum;
}

int main()
{
    int t, cnt = 0;
    scanf("%d", &t);
    while(t--)
    {
        printf("Case %d:", ++cnt);
        scanf("%d%d", &n, &q);
        for (int i = 0;i < n;i++)
            scanf("%lld", &A[i]);
        LL sum = 0;
        for (int i = 0;i < n;i++)
        {
            sum += (n-1-i)*A[i];
            sum -= i*A[i];
        }
        int op, x, v;
        puts("");
        while(q--)
        {
            scanf("%d", &op);
            if (op == 0)
            {
                scanf("%d%d", &x, &v);
                sum -= (n-1-x)*A[x];
                sum += x*A[x];
                A[x] = v;
                sum += (n-1-x)*A[x];
                sum -= x*A[x];
            }
            else
            {
                printf("%lld\n", sum);
            }
        }
    }

    return 0;
}