【C++】PAT（advanced level）1046. Shortest Distance (20)

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ... 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 (<=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 14 931 32 54 1

Sample Output:

3107

1.好吧，总数减去部分的和，再比大小

#include <iostream>using namespace std;int Min(int d1,int d2){if(d1>d2){return d2;}else{return d1;}}int main(){int n;while(cin>>n){int i,sum=0;int *dis=new int[n+1];dis[0]=0;for(i=0;i<n;i++){int distance;cin>>distance;sum+=distance;dis[i+1]=sum;}int m;cin>>m;for(i=0;i<m;i++){int a,b;cin>>a>>b;if(a>b){int d1=dis[a-1]-dis[b-1];int d2=sum-d1;cout<<Min(d1,d2)<<endl;}else{int d1=dis[b-1]-dis[a-1];int d2=sum-d1;cout<<Min(d1,d2)<<endl;}}}return 0;}