Time Limit: 3000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
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
1
3 5
1 2 3
1
0 0 3
1
0 2 1
1
Sample Output
Case 1:
-4
0
4
Source
#include#define N 100000+100typedef long long LL;LL a[N];int main(){ int t, Q=1; scanf("%d", &t); while(t--) { int n, q; scanf("%d%d", &n, &q); for(int i=0; i n) continue; getSum +=A*a[i]-b*a[i]; A--; b++; } int firNum; printf("Case %d:\n", Q++); while(q--) { scanf("%d", &firNum); if(firNum==1) printf("%lld\n", getSum); else { int posi; LL modify; scanf("%d%lld", &posi, &modify); LL wc=modify-a[posi]; getSum=getSum+(n-1-2*posi)*wc; a[posi]=modify; //更新必须要有 !!!!! ; } } } return 0;}