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 N integer 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
代码
#include<stdio.h>
#include<string.h>
int main()
{
int n,m;
int qi,zhi;
int s=0;
scanf("%d “,&n);
int dis[n];
int d[n];
int tmp;
int i,j;
int k;
int juli[n];
memset(dis,0,sizeof(dis));
memset(d,0,sizeof(d));
memset(juli,0,sizeof(juli));
for(i=0;i<n;i++)
{
scanf(”%d",&d[i]);
dis[i]=s;
s+=d[i];
}
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%d%d",&qi,&zhi);
if(zhi<qi)
{
tmp=zhi;
zhi=qi;;
qi=tmp;
}
juli[i]=dis[zhi-1]-dis[qi-1];
if(s-juli[i]<juli[i])
juli[i]=s-juli[i];
}
for(k=0;k<m;k++)
{
printf("%d\n",juli[k]);
}
return 0;
}