1046 Shortest Distance (20 分)

PAT原题

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, . . . , where D_i​​ is the distance between the i-th and the (i+1)-st exits, and D_Nis 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

暴力解答：

#include 
using namespace std;
const int maxn = 100005;
int a[maxn];
int b[maxn];
int n, m;
int instance(int x, int y) {
    int z=0, f=0; //例如x=1, y=3: z表示例如1->2->3的距离，f表示从 3->...>n 1->3 距离
    if (x > y) {
        int t;
        t = x;
        x = y;
        y = t;
    }
    for (int i = x; i < y; i++) {
        z += a[i];
    }
    for (int i = y; i <= n; i++) {
        f += a[i];
    }
    for (int i = 1; i< x; i++) {
        f += a[i];
    }
    return z>f?f:z;
}
int main() {
	freopen("data.txt","r",stdin);
    int x, y;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    scanf("%d", &m);
    int short_instance;
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &x, &y);
        short_instance = instance(x, y);
        printf("%d\n", short_instance);
    }
	return 0;
}