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 [3,105]), followed by N integer distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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 (≤104), 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 107.
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
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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;
}
}