【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 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

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;
}