1046. Shortest Distance (20)【贪心】——PAT (Advanced Level) Practise

题目信息

1046. Shortest Distance (20)

时间限制100 ms

内存限制65536 kB

代码长度限制16000 B

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 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 (<=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

解题思路

从前往后和从后往前（过最后一个再过第一个）取最小

AC代码

#include <cstdio>
#include <algorithm>
using namespace std;
long long a[100005];
int main()
{
int n, m, x, y;
scanf("%d", &n);
for (int i = 1; i <= n; ++i){
scanf("%lld", a + i);
a[i] += a[i-1];
}
scanf("%d", &m);
while (m--){
scanf("%d%d", &x, &y);
if (y < x) {
swap(x, y);
}
printf("%d\n", min(a[y-1] - a[x-1], a[x-1] + a
- a[y-1]));
}
return 0;
}