Not Only Algorithm,不仅仅是算法,关注数学、算法、数据结构、程序员笔试面试以及一切涉及计算机编程之美的内容 。。
你的位置:NoAlGo博客 » 题解 » 

PAT 1046. Shortest Distance

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

解答

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const int mx = 100000 + 10;
int a[mx] = {0}, n, total = 0;

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", a+i), total += a[i], a[i] += a[i-1];

	int m; scanf("%d", &m);
	while (m--)
	{
		int x, y; scanf("%d%d", &x, &y); x--, y--;
		printf("%d\n", min(abs(a[x]-a[y]), total-abs(a[x]-a[y])));
	}
	return 0;
}
上一篇: 下一篇:

我的博客

NoAlGo头像编程这件小事牵扯到太多的知识,很容易知其然而不知其所以然,但真正了不起的程序员对自己程序的每一个字节都了如指掌,要立足基础理论,努力提升自我的专业修养。

站内搜索

最新评论