A1046 Shortest Distance (20分)_the task is really simple: given n exits on a…-优快云博客

本文链接：https://blog.youkuaiyun.com/WeDon_t/article/details/105150943

本文介绍了一种计算环形高速公路上任意两点间最短距离的算法。通过正序计算总距离并减去两出口间的直接距离，比较得出最小值。此算法适用于输入点构成环形的情况。

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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 D1D2 ⋯ DN, where Diis 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 7), 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

题目大意：

输入的各个点围成一个环，其中相邻两个点之间的距离为输入的值
计算两个点最小的距离

思路：

先正序计算，总体减去，进行比较

#include <stdio.h>
const int MAXN=100010;
int array[MAXN];
int array1[MAXN];
void swap(int &a,int &b){
	a=a^b;
	b=a^b;
	a=a^b;
}
int main(){
	int n;
	scanf("%d",&n);
	
	int s=0;
	for(int i=1;i<=n;i++){
		scanf("%d",&array[i]);
		s+=array[i];
		array1[i]=s;
	}
	int m;
	scanf("%d",&m);
	while(m--!=0){
		int a,b;	
		scanf("%d%d",&a,&b);
		if(a>b)swap(a,b);
		int sum=0;
		sum = array1[b-1]-array1[a-1];
		sum = sum<s-sum?sum:s-sum;
		printf("%d\n",sum);
		
	}
}