## PAT——A1046 原

Cinzano

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```

``````#include<stdio.h>
int main(){
int N,a[100000]={0},t,count=0;
scanf("%d",&N);//N表示即将要输入的exits数；
for(int i=2;i<=N;i++){
scanf("%d",&t);//t表示即将输入的每两个exits之间的distance；
a[i]=a[i-1]+t;//a[i]表示1-st exit to i-th 的 total distance;
}
scanf("%d",&t);//this input is the last in N, this 't' is the distance of N-th to 1-st;
count=a[N]+t;//count equal 前N个出口总距离 + 最后一个输入的第N个exit到第一个exit的距离；
int M,x,y,m=0;
scanf("%d",&M);
while(M--){
scanf("%d %d",&x,&y);
if(x>y){
int temp=x;
x=y;
y=temp;
}
m=a[y]-a[x];//m is the distance of x-th to y-th;
int min=m<(count-m)? m:(count-m);//short distance need comparing;
printf("%d\n",min);
m=0;
}
}``````

© 著作权归作者所有

### Cinzano

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