1046. Shortest Distance (20)-PAT甲级

题目：

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

这道题画个图会理解的很清楚，需要注意的是时间要求是100ms，所以说
一定要注意尽量不要让计算重复，怎么做那？
就是把它存起来，用空间换取时间。答案中就是用dis数组来储存距离的和
计算的时候只要一个减法就够了。
为了让计算统一，我们可以按顺时针来计算，要求最小值只要取得”,min(temp,sum-temp))就行了。

解答：

#include<cstdio>#include<algorithm>using namespace std;const int MAXN=100005;int dis[MAXN],A[MAXN];//A[i]表示i到i+1位置的距离 int main(){    int sum=0,n,left,right;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d",&A[i]);        sum+=A[i];        dis[i]=sum;//dis[i]预存1到i位置的距离之和，减少了以后的计算     }    int query;    scanf("%d",&query);    while(query--){        scanf("%d%d",&left,&right);        if(left>right)swap(left,right);        int temp=dis[right-1]-dis[left-1];        printf("%d\n",min(temp,sum-temp));//一定有两条线路，取其中的最小距离     }     return 0;}

2017/8/23添加

#include<cstdio>#include<vector>using namespace std;int main(){    int n,m;    scanf("%d",&n);    vector<int> d(n+1);    for(int i=1;i<=n;i++){        scanf("%d",&d[i]);        d[i]+=d[i-1];    }    scanf("%d",&m);    while(m--){        int ans,a,b;        scanf("%d%d",&a,&b);        if(a>b)swap(a,b);        ans=d[b-1]-d[a-1];        printf("%d\n",min(d[n]-ans,ans));    }    return 0;}