PAT甲 1046. Shortest Distance (20) 2016-09-09 23:17 22人阅读 评论(0) 收藏

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ... D_N,
where D_i is the distance between the i-th and the (i+1)-st exits, and D_N 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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
	int a[100005],n,m,l,r,x,y,k,sum;
	scanf(" %d",&n);
	a[0]=0;
	sum=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&k);
		a[i]=a[i-1]+k;
		sum+=k;
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&l,&r);
		if(l>r)
			swap(l,r);
		x=a[r-1]-a[l-1];
		y=sum-x;

		printf("%d\n",(x<y)?x:y);
	}
	return 0;
}