1369 - Answering Queries（规律）

1369 - Answering Queries

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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 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+100;
typedef long long LL;
LL a[MAXN],b[MAXN];
int main(){
    int T,n,q,cnt=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&q);
        LL sum=0;
        for(int i=0;i<n;i++)scanf("%lld",a+i);
        sum=0;
        for(int i=n-1;i>=1;i--)sum+=a[i],b[i-1]=sum;
                //for(int i=0;i<n;i++)printf("%d ",b[i]);puts("");
        b[n-1]=0;
        sum=0;
        for(int i=0;i<n-1;i++)sum+=(a[i]*(n-i-1)-b[i]);
        mem(b,0);
        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-1)-x*(v-a[x]);
                a[x]=v;
            }
        }
    }
    return 0;
}