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| 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 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.
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 |
Dataset is huge, use faster I/O methods.
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<math.h> 6 #include<stdlib.h> 7 typedef long long LL; 8 LL bns[200000]; 9 int main(void) 10 { 11 int k; 12 int i,j; 13 scanf("%d",&k); 14 int s; 15 int p,q; 16 LL ans=0; 17 for(s=1; s<=k; s++) 18 { 19 scanf("%d %d",&p,&q); 20 for(i=1; i<=p; i++) 21 { 22 scanf("%lld",&bns[i]); 23 } 24 ans=0; 25 for(i=1; i<=p; i++) 26 { 27 ans+=(LL)(p-2*i+1)*(LL)bns[i]; 28 } 29 printf("Case %d:\n",s); 30 while(q--) 31 { 32 int ask; 33 int n,m; 34 scanf("%d",&ask); 35 if(ask==1) 36 { 37 printf("%lld\n",ans); 38 } 39 else 40 { 41 scanf("%d %d",&n,&m); 42 43 { 44 ans-=(LL)(p-2*(n+1)+1)*bns[n+1]; 45 ans+=(LL)(p-2*(n+1)+1)*(LL)m; 46 bns[n+1]=m; 47 } 48 } 49 } 50 } 51 return 0; 52 }
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原文地址：http://www.cnblogs.com/zzuli2sjy/p/5539158.html
