1046 Shortest Distance(20 分)
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
解题 思路:这题比较简单,直接看代码就懂了。
#include<bits/stdc++.h>
using namespace std;
int a[100010],dis[100010];
int main(void)
{
int n;
scanf("%d",&n);
memset(dis,0,sizeof dis);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
dis[i+1]=dis[i]+a[i];
}
int sum=dis[n+1];
int m;
scanf("%d",&m);
while(m--)
{
int l,r;
scanf("%d%d",&l,&r);
if(l>r) swap(l,r);
int temp=dis[r]-dis[l];
printf("%d\n",min(temp,sum-temp));
}
return 0;
}