pat A1046

1046. Shortest Distance (20)

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 ... 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 (<=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

错误代码

#include <iostream>
#include <vector>
using namespace std;


int main(void)
{
int n, m;
int total = 0;
vector<int> A;
cin >> n;
int ival;
for (int i = 0; i < n; ++i)
{
cin >> ival;
A.push_back(ival);
total += A[i];
}
cin >> m;
int a, b;

for (int i = 0; i < m; ++i)
{
cin >> a >> b;

if (a > b)
{
int tmp = a;
a = b;
b = tmp;


}
int s1 = 0, s2 = 0;
for (int j = a; j < b; ++j)
{
s1 += A[j - 1];
}
s2 = total - s1;
s1 = s1 > s2 ? s2 : s1;
cout << s1 << endl;
}
return 0;
}

.错误原因：超时了

错在两层for循环处。

解决办法设置dis【】数组，对该数组预处理，就可减去一层for循环。

代码：

#include <iostream>
#include <vector>
using namespace std;


int main(void)
{
int n, m;
int total = 0;

vector<int> A;
cin >> n;
vector<int> dis;

int ival;
for (int i = 0; i < n; ++i)
{
cin >> ival;
A.push_back(ival);
total += A[i];
dis.push_back(total);
}
cin >> m;
int a, b;

for (int i = 0; i < m; ++i)
{
cin >> a >> b;

if (a > b)
{
int tmp = a;
a = b;
b = tmp;


}
int s1 = 0;
if (a < 2)
s1 = dis[b - 2];
else
s1 = dis[b - 2] - dis[a - 2];
int s2 = total - s1;
s1 = s1 > s2 ? s2 : s1;
cout << s1 << endl;
}
return 0;
}