A1046 Shortest Distance [简单模拟--卡算法复杂度]

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ D​N​​, where D​i​​ is the distance between the i-th and the (i+1)-st exits, and D​N​​ is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (≤10​4​​), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10​7​​.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1
Sample Output:

3
10
7

————————————————
这个第三个测试点，考察复杂度，卡时间点，原本的复杂度太高了，达到了M*N的复杂度，采用走一步存一步的方法，这样复杂度讲到了常数1。

#include<iostream>
using namespace std;
int main()
{
	int n, m, a[100001] = { 0 }; int sum = 0;int dis[100001];
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
        sum += a[i];
        dis[i]=sum;
	}
	cin >> m; int c, d; 
	for (int i = 0; i < m; i++)
	{
		int temp1 = 0;
		cin >> c >> d;
		if (c > d)
		{
			int temp = c;
			c = d;
			d = temp;
		}
        temp1=dis[d-1]-dis[c-1];
		int temp = sum - temp1;
		if (temp <= temp1)
			cout << temp << endl;
		else
			cout << temp1<< endl;
	}
}

这个是原本的

#include<iostream>
using namespace std;
int main()
{
	int n, m, a[11110] = { 0 }; int sum = 0;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
        sum += a[i];
	}
	cin >> m; int c, d; 
	for (int i = 0; i < m; i++)
	{
		int dis = 0;
		cin >> c >> d;
		if (c > d)
		{
			int temp = c;
			c = d;
			d = temp;
		}
		for (int i = c; i < d; i++)
		{
			dis += a[i];
		}
		int temp = sum - dis;
		if (temp <= dis)
			cout << temp << endl;
		else
			cout << dis << endl;
	}
}