【PAT 甲級】1046 Shortest Distance (20)

A1046 Shortest Distance (20)

 

題目描述

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.

輸入描述

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.

輸出描述

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

輸入例子

5 1 2 4 14 9
3
1 3
2 5
4 1

輸出例子

3
10
7

我的思路

 反向距離=總距離-正向距離，最小距離是反或正向距離。

 

我的代碼

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
	long int N, M, a, b, distance=0, sum=0;
	scanf("%d", &N);
	int *d = new int[N+2];
	
	for (int i=1; i<=N; i++)
	{
	    scanf("%d", &d[i]);
	    sum += d[i];
	}
	scanf("%d", &M);
	int *minD = new int[M+2];
	for (int i=1; i<=M; i++)
	{
		distance = 0;
	    scanf("%d %d", &a, &b);
	    if (a>b)
		{  //始終保持a<b 
	    	int temp = a;
	    	a = b;
	    	b = temp;
		}
	    for (int j=a; j<b; j++)
		{
	        distance += d[j];	
		}
		minD[i] = min(distance, sum-distance);  //反向距離=總距離-正向距離 
	}
		
	for(int i=1; i<=M; i++)
	{
		printf("%d", minD[i]);
		if (i!=M) 
		{  //輸出格式 
			printf("\n");
		}
	}
	return 0;
}