lightoj 1369 - Answering Queries 【思维】
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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
1
3 5
1 2 3
1
0 0 3
1
0 2 1
1
Case 1:
-4
0
4
Note
Dataset is huge, use faster I/O methods.
#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN 100000+10#define MAXM 50000000#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;LL A[MAXN];int main(){ int t, kcase = 1; Ri(t); W(t) { int n, q; Ri(n); Ri(q); LL sum = 0; for(int i = 0; i < n; i++) { Rl(A[i]); sum += A[i]; } sum -= A[0]; LL ans = 0; for(int i = 0; i < n-1; i++) { ans += A[i] * (n-i-1) - sum; sum -= A[i+1]; } printf("Case %d:\n", kcase++); while(q--) { int op; Ri(op); if(op == 1) Pl(ans); else { int x, v; Ri(x); Ri(v); ans -= (v-A[x]) * x + (A[x]-v) * (n-x-1); A[x] = v; } } } return 0;}
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