1369

1369 - Answering Queries

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

3 5

1 2 3

0 0 3

0 2 1

Case 1:

-4

题解：我的思路本来是针对每次的修改，都在询问里面找值，不出意外肯定超时了，出来看了大神的题解，是针对每次修改再修改sum就妥了，比赛的时候就没想到。。。

代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define mem(x,y) memset(x,y,sizeof(x))
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 const int MAXN=1e5+100;
10 typedef long long LL;
11 LL a[MAXN],b[MAXN];
12 int main(){
13     int T,n,q,cnt=0;
14     scanf("%d",&T);
15     while(T--){
16         scanf("%d%d",&n,&q);
17         LL sum=0;
18         for(int i=0;i<n;i++)scanf("%lld",a+i);
19         sum=0;
20         for(int i=n-1;i>=1;i--)sum+=a[i],b[i-1]=sum;
21                 //for(int i=0;i<n;i++)printf("%d ",b[i]);puts("");
22         b[n-1]=0;
23         sum=0;
24         for(int i=0;i<n-1;i++)sum+=(a[i]*(n-i-1)-b[i]);
25         mem(b,0);
26         printf("Case %d:
",++cnt);
27         while(q--){
28             int t,x,v;
29             scanf("%d",&t);
30             if(t){
31                 printf("%lld
",sum);
32                 }
33             else{
34                 scanf("%d%d",&x,&v);
35                 sum=sum+(v-a[x])*(n-x-1)-x*(v-a[x]);
36                 a[x]=v;
37             }
38         }
39     }
40     return 0;
41 }