PAT 1046 Shortest Distance（前缀和）

1046 Shortest Distance

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

总结：因为道路是一个环，所以两个点之间的距离有两种计算方法（都只能是从小点往大点走），从右点到左点，从左点到右点，两段距离相加就是每两个点之间距离的总和，所以只需要计算其中一段就可以直到另一段距离的大小了，选择其中较小的一段即为最后的答案

#include <iostream>
using namespace std;

int main(){
    int n,m,l,r;
    cin >> n;
    int a[n]={0};
    
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        a[i]+=a[i-1];
    }
   
    cin >> m;
    for(int i=0;i<m;i++){
        int l,r,s1,s2;
        scanf("%d%d",&l,&r);
        if(l>r) swap(l,r);
        s1=a[r-1]-a[l-1];
        s2=a[n]-s1;
        printf("%d\n",min(s1,s2));
    }
    
    return 0;
}

 好好学习，天天向上！

我要考研！

2022.11.10

//前缀和
#include <iostream>
using namespace std;

const int N=100010;
int a[N];
int main(){
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        a[i]+=a[i-1];
    }
    int total=a[n];
    scanf("%d",&m);
    for(int i=0;i<m;i++){
        int l,r;
        scanf("%d%d",&l,&r);
        if(l>r) swap(l,r);
        //反方向：前缀和      顺方向：总的路程-反方向的路程
        printf("%d\n",min(a[r-1]-a[l-1],total-(a[r-1]-a[l-1])));
    }
    
    return 0;
}