1046 Shortest Distance

1046 Shortest Distance (20 point(s))

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 Ninteger distances D​1​​ D​2​​ ⋯ 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 (≤10​4​​), 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

简单模拟。

常规方法会超时（case#2不能通过，17分），用hashTable初始化sum[]，记录从1到i+1的元素和，为后面的每次询问准备，减小复杂度。 

TLE总结：1. 采用scanf、printf输入输出；2. 空间换时间。特别是查询问题（本题）、素数筛法等。

#include<iostream>
using namespace std;
const int N = 1e5+7; 
int num[N]={0};int sum[N]={0};int n;
void init(){
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+num[i];//sum[i] 从1到i+1的距离 
}
int shortdis(int p1,int p2){
	if(p1==p2) return 0;
	if(p1>p2){
		int tmp = p1;p1 = p2;p2 = tmp;
	}
	int a1 = sum[p2-1]-sum[p1-1];
	int a2 = sum[n]-a1;
	return a1>a2?a2:a1;
}
int main(void){
	int m,p1,p2;cin>>n;
	for(int i=1;i<=n;i++) cin>>num[i];
	init();
	cin>>m;
	while(m--){
		cin>>p1>>p2;
		int ans = shortdis(p1,p2);
		cout<<ans<<endl;
	}
	return 0;
}