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 D1D2 ⋯ DN, where Diis 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 (≤10 7), 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
题目大意:
- 输入的各个点围成一个环,其中相邻两个点之间的距离为输入的值
- 计算两个点最小的距离
思路:
- 先正序计算,总体减去,进行比较
#include <stdio.h>
const int MAXN=100010;
int array[MAXN];
int array1[MAXN];
void swap(int &a,int &b){
a=a^b;
b=a^b;
a=a^b;
}
int main(){
int n;
scanf("%d",&n);
int s=0;
for(int i=1;i<=n;i++){
scanf("%d",&array[i]);
s+=array[i];
array1[i]=s;
}
int m;
scanf("%d",&m);
while(m--!=0){
int a,b;
scanf("%d%d",&a,&b);
if(a>b)swap(a,b);
int sum=0;
sum = array1[b-1]-array1[a-1];
sum = sum<s-sum?sum:s-sum;
printf("%d\n",sum);
}
}