﻿PAT(A) 1046. Shortest Distance (20)-Ocrosoft

# PAT(A) 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 … 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
```

```#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define strend string::npos
#define ms(a) memset(a,0,sizeof(a))
#define rep(a,v,b) for(int a=v;a<b;a++)
#define repe(a,v,b) for(int a=v;a<=b;a++)
#define pre(a,v,b) for(int a=v;a>b;a--)
#define pree(a,v,b) for(int a=v;a>=b;a--)
#define lowbit(x) x&-x
typedef long long LL;
const LL LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const int MAXN = 1e5 + 10;
const int MOD = 1000000007;
int gcd(int a, int b)
{
if (!b)return a;
return gcd(b, a%b);
}
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
/*(◕‿‿◕) 签订契约,成为马猴烧酒吧 (◕‿‿◕)*/
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
using namespace std;
int n;
int a[MAXN], b[MAXN] = { 0 };
int ans = INF;
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
b[i] = b[i - 1] + a[i];
}
int m; cin >> m;
while (m--)
{
int x, y; scanf("%d%d", &x, &y);
if (x > y)swap(x, y);
ans = (b[y-1] - b[x-1]);
ans = min(ans, (b[n] - b[y - 1] + b[x - 1]));
printf("%d\n", ans);
}
return 0;
}
```