Taking Bus

                                                              Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768
K (Java/Others)

Problem Description

Bestland has a very long road and there are n bus
station along the road, which are numbered 1 to n from
left to right. There are m persons
wanting to take the bus to some other station. You task is to find the time needed for each person. Note: All the other information you need is below. Please read the statment carefully.

Input

There are multiple test cases. The first line of input contains an integer 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

For each person, you should output one integer which is the minimum time needed before arriving bus station yi.

Sample Input

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

Sample Output

21
10
28

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

四种情况考虑一下就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100000+100;
long long sum[maxn];
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        long long x;
        scanf("%d%d",&n,&m);
        sum[1]=0;
        for(int i=2;i<=n;i++)
        {
            scanf("%I64d",&x);
            sum[i]=sum[i-1]+x;
        }
        int u,v;
        long long ans;
        for(int w=1;w<=m;w++)
        {
            int i=((w-1)%n)+1;
            scanf("%d%d",&u,&v);
            if(u>=i)
            {
                if(v>u)
                {
                    ans=sum[v]-sum[i];
                }
                else
                {
                    ans=sum[n]-sum[i]+sum[n]-sum[v];
                }
            }
            else
            {
                if(v>u)
                {
                    ans=sum[n]-sum[i]+sum[n]+sum[v];
                }
                else
                {
                    ans=sum[n]-sum[i]+sum[n]-sum[v];
                }
            }
            printf("%I64d\n",ans);
        }
    }
    return 0;
}