lightoj 1369 - Answering Queries （数学、规律、水）

Time Limit:3000MS    Memory Limit:32768KB    64bit IO Format:%lld & %llu（明确指出64位格式为lld，用I64d会wa！！！）

Description

The problem you need to solve here is pretty simple. You are give a functionf(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 integern, 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 findf 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 containsn 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

1

3 5

1 2 3

1

0 0 3

1

0 2 1

1

Sample Output

Case 1:

-4

0

4

这道题吧，其实水水就过去了，刚开始以为要用线段树来做，后来发现了个求和规律= =，也就是，假设有n个数，那么根据上面的代码求出的sum方法，可推出一个求和规律，由于下标是从0开始，所以先将n减一，然后可得求和规律即sum=（n）*a[0]+(n-2)*a[1]+(n-4)*a[2]+(n-6)*a[3]+(n-8)*a[4]+……(n-2*n)*a[n-1]；也就是说，他的和等于系数从n-1开始每次减2，乘上a[i]；（不理解的可以验证几组数据很容易就发现了），求和公式得出来了，那么如果进行更新时候，只需要将要更新的一项去掉，然后加上该项更新后的值就行了，如果进行询问，直接输出sum；

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#define N 50010#define INF 999999999using namespace std;int a[100010],b[100010];int main(){    int n,p,q,i,x,v,h;    long long sum;//注意用long long不然会溢出；    int t;    int k=0;    scanf("%d",&t);    while(t--)    {        sum=0;        scanf("%d%d",&n,&q);        for(i=0;i<n;i++)            scanf("%d",&a[i]);        h=n;        n--;        for(i=0; i<h; i++)        {            sum+=((long long)n)*a[i];//进行求和运算；            b[i]=n;把每一项对应的系数存起来，方便后面的更新；            n=n-2;        }        printf("Case %d:\n",++k);        while(q--)        {            scanf("%d",&p);            if(p==0)            {                scanf("%d%d",&x,&v);                sum-=((long long)b[x])*a[x];//如果要把a[x]的值更新为v，那么先将这一项与对应系数的乘积减去；                sum+=((long long)b[x])*v;//然后再加上更新后的值与对应系数的乘积；                a[x]=v;//这里很重要，注意把a[x]更新为v            }            if(p==1)                printf("%lld\n",sum);//输出用lld，lld，不能用I64d，因为题目上明确说了64位输出格式为lld，我只想说没看见题目前面的要求，用I64wa了N次= =        }    }    return 0;}