R - Answering Queries

R - Answering Queries

Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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Status
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

Hint
Dataset is huge, use faster I/O methods.

O(1) per query

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1. 题意：给出一个数列，让你根据题意推导出公式，然后根据选择输出所需信息。
2. 思路：将所给的循环化简成人类能看得懂的语言，其实就是加加减减。
3. 失误：没有将a[i]换成v,忽略了他对后面输出的影响，还有输出格式，应该是输入一个命令，输出一个数据，不是将所有一起输出。
4. 代码如下：

 #include<cstdio>long long a[100000+10];int main(){    long long  t,n,s,flag,x,v,k=0,i,q,j;    scanf("%lld",&t);    while(t--)    {        scanf("%lld %lld",&n,&q);//输入时 格式控制符容易错 并且不易查找 应注意         s=0;        for(i=0;i<n;++i)        {            scanf("%lld",&a[i]);            s+=(n-1)*a[i]-2*i*a[i];//推出公式         }        printf("Case %lld:\n",++k);        for(i=1;i<=q;++i)        {            scanf("%lld",&flag);            if(flag==1)            {                  printf("%lld\n",s);              }            else            {                scanf("%lld %lld",&x,&v);                s=s+(n-1)*v-2*x*v-(n-1)*a[x]+2*x*a[x];                a[x]=v;//少写一步，逻辑还是不够呵             }        }    }    return 0; }