LightOJ 1369 - Answering Queries(规律)

[align=center]1369 - Answering Queries[/align]

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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

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 and106 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

题意就不说了，按照题目给的函数写，果断超时，于是想了一会发现sum+=a[i]*(n-1-i*2); 于是两层for循环改成了一次， 以为会过，又超时了。QAQ。。。  后来看队友的写法，当数组中值改变时，直接改sum的值，那样从头到尾只用了一次for循环求值。

当i=0时，有a0-a1+a0-a2+a0-a3+.....，当i=1时，有a1-a2+a1-a3+a1+.....,观察可得发现第一个式子刚好比第二个多出来 n-i 个（a0-a1)。之后操作时要改变值时直接在之前求得的总和身上改，例如改变了 a7 增加了2 则从第8到n 个数被 a7减时结果都要多2，而从第0个到第6个减 a7 时结果都小2 。

具体代码如下：

#include<cstdio>
#include<cstring>
#define maxn 100010
int a[maxn];

long long f(int a[],int n)
{
long long sum=0,temp=0;
int i;
for(i=1;i<n;++i)
temp+=a[0]-a[i];
sum=temp;
for(i=1;i<n;++i)
{
long long cnt=a[i]-a[i-1];//此处必须将a[i]-a[i-1]的结果转化为long long 型。
temp+=(n-i)*cnt;
sum+=temp;
}
return sum;
}

int main()
{
int t,n,x,v,q,k=1,i,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&q);
for(i=0;i<n;++i)
scanf("%d",&a[i]);
long long sum=f(a,n);
printf("Case %d:\n",k++);
while(q--)
{
scanf("%d",&m);
if(m)
printf("%lld\n",sum);
else
{
scanf("%d%d",&x,&v);
long long temp=v-a[x];
a[x]=v;
sum+=(n-1-x-x)*temp;
}
}
}
return 0;
}