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 nis 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
1
3 5
1 2 3
1
0 0 3
1
0 2 1
1
Sample Output
Case 1:
-4
0
4
Note
Dataset is huge, use faster I/O methods.
有两个操作:
0 x v将A[x]转变成v。
1计算函数f的值
分析:
推出规律
n个数,用a,b,c......表示 ,推出来为:
(n-1)*a[1]+(n-3)*a[2]+(n-5)*a[3].......
归纳话的规律为:
(n-(2*i-1))*a[i]
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
const int N=1e5+5;
int a[N];
int main(){
int T;
scanf("%d",&T);
int cas=0;
while(T--){
int n,q;
scanf("%d%d",&n,&q);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
printf("Case %d:\n",++cas);
LL sum=0;
for(int i=0;i<n;i++){
sum+=(LL)(n-1-2*i)*a[i];
}
while(q--){
int op;
scanf("%d",&op);
if(op==1){
printf("%lld\n",sum);
}
else{
int pos,v;
scanf("%d%d",&pos,&v);
sum-=(LL)(n-2*pos-1)*a[pos];
sum+=(LL)(n-2*pos-1)*v;
a[pos]=v;
}
}
}
}