PAT_A1046 1046 Shortest Distance (20 分)

題目

• 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 $[3,10 ​^5]$), 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

• 源碼參考：

/*
myarr[i]存放從頭到i處的距離，myarr[0]存放0；
mydis[i]在開始時存放輸入的距離數組，後面用作存放輸出的兩點間距離；
sum存放距離之和。
*/
#include <cstdio>
#include <iostream>
using namespace std;

const int maxn=100010;
int myarr[maxn],mydis[maxn];
int sum=0;

int main()
{
    int N;
    scanf("%d",&N);
    for(int i=0;i<N;++i)
    {
        scanf("%d",&mydis[i]);
        sum=sum+mydis[i];
        if(i==0)
        {
           myarr[i]=0;
        }
        else
        {
            myarr[i]=myarr[i-1]+mydis[i-1];
        }
    }
    int M;
    scanf("%d",&M);
    int a,b;
    for(int i=0;i<M;++i)
    {
        scanf("%d %d",&a,&b);
        if(a>b)
        {
            mydis[i]=myarr[a-1]-myarr[b-1];
        }
        else
        {
            mydis[i]=myarr[b-1]-myarr[a-1];
        }
        if(mydis[i]>sum-mydis[i])
        {
            mydis[i]=sum-mydis[i];
        }
    }
    for(int i=0;i<M;++i)
    {
        printf("%d\n",mydis[i]);
    }
    return 0;
}