1369 - Answering Queries(规律)
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 |
1 3 5 1 2 3 1 0 0 3 1 0 2 1 1 |
Case 1: -4 0 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:
- #include<iostream>
- #include<cstdio>
- #include<cstring>
- #include<cmath>
- #include<algorithm>
- #define mem(x,y) memset(x,y,sizeof(x))
- using namespace std;
- const int INF=0x3f3f3f3f;
- const int MAXN=1e5+;
- typedef long long LL;
- LL a[MAXN],b[MAXN];
- int main(){
- int T,n,q,cnt=;
- scanf("%d",&T);
- while(T--){
- scanf("%d%d",&n,&q);
- LL sum=;
- for(int i=;i<n;i++)scanf("%lld",a+i);
- sum=;
- for(int i=n-;i>=;i--)sum+=a[i],b[i-]=sum;
- //for(int i=0;i<n;i++)printf("%d ",b[i]);puts("");
- b[n-]=;
- sum=;
- for(int i=;i<n-;i++)sum+=(a[i]*(n-i-)-b[i]);
- mem(b,);
- printf("Case %d:\n",++cnt);
- while(q--){
- int t,x,v;
- scanf("%d",&t);
- if(t){
- printf("%lld\n",sum);
- }
- else{
- scanf("%d%d",&x,&v);
- sum=sum+(v-a[x])*(n-x-)-x*(v-a[x]);
- a[x]=v;
- }
- }
- }
- return ;
- }
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