hdu5163---Taking Bus

Taking Bus

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 100    Accepted Submission(s): 55

Problem Description

n bus station along the road, which are numbered 1 to n from left to right. There are m

 

Input

T (1≤T≤60), indicating the number of test cases. For each test case: The first line contains two integers n and m (2≤n,m≤105), indicating the number of bus stations and number of people. In the next line, there are n−1 integers, d1,d2,…,dn−1 ( 1≤di≤109). The i-th integer means the distance between bus station i and i+1 is di ( 1≤i<n). In the next m lines, each contains two integers xi and yi ( 1≤xi,yi≤n,xi≠yi), which means i-th person is in bus station xi and wants goto bus station yi. (1≤i≤m)

What else you should know is that for the i-th person, the bus starts at bus station ((i−1) mod n)+1 and drives to right. When the bus arrives at station n, it will turn around and drive from right to left. Similarly, When the bus arrives at station 1, it will turn around and drive from left to right. You can assume that the bus drives one meter per second. And you should only consider the time that the bus drives and ignore the others.

 

Output

yi.

 

Sample Input

1
7 3
2 3 4 3 4 5
1 7
4 5
5 4

 

Sample Output

Hint

For the first person, the bus starts at bus station 1, and the person takes in bus at time 0. After 21 seconds, the bus arrives at bus station 7. So the time needed is 21 seconds. For the second person, the bus starts at bus station 2. After 7 seconds, the bus arrives at bus station 4 and the person takes in the bus. After 3 seconds, the bus arrives at bus station 5. So the time needed is 10 seconds. For the third person, the bus starts at bus station 3. After 7 seconds, the bus arrives at bus station 5 and the person takes in the bus. After 9 seconds, the bus arrives at bus station 7 and the bus turns around. After 12 seconds, the bus arrives at bus station 4. So the time needed is 28 seconds.

 

Source

BestCoder Round #27

 

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处理一个前缀和就行

/*************************************************************************
    > File Name: bc27b.cpp
    > Author: ALex
    > Created Time: 2015年01月24日 星期六 19时11分51秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;
const int N = 200010;
struct node
{
	int s, e;
}peo[N];

LL xis[N];
LL dist[N];

int main ()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int s, e;
		int n, m;
		scanf("%d%d", &n, &m);
		memset (xis, 0, sizeof(xis));
		for (int i = 1; i < n; ++i)
		{
			scanf("%I64d", &dist[i]);
		}
		xis[1] = 0;
		for (int i = n; i < 2 * n - 2; ++i)
		{
			dist[i] = dist[2 * n - i - 2];
		}
		for (int i = 2; i < 2 * n - 1; ++i)
		{
			xis[i] = xis[i - 1] + dist[i - 1];
		}
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d%d", &s, &e);
			int start = (i - 1) % n + 1;
			if (s <= e)
			{
				if (start <= s)
				{
					printf("%I64d\n", xis[e] - xis[start]);
				}
				else
				{
					printf("%I64d\n", xis[n] - xis[start] + xis[n] + xis[e]);
				}
			}
			else
			{
				printf("%I64d\n", xis[n] - xis[start] + xis[n] - xis[e]);
			}
		}
	}
	return 0;
}