PAT 1046. Shortest Distance (20)（简单加法）

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1046. Shortest Distance (20)

时间限制
100 ms
内存限制
65536 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 … 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

题目大意

• 1.给出一个环和每个节点的距离，求环中俩点最短的距离，即问你是往那个方向走。

AC代码

#include<iostream>
#include<math.h>

using namespace std;
int p[100001],sum[100001];
int main(int argc, char *argv[])
{
    int n,m;sum[0] = 0;
    cin >> n;
    //sum[0] = 0;
    for (int i = 1; i <= n; ++i) {
        cin >> p[i];
        sum[i] = sum[i-1] +  p[i];
    }
    cin >> m;
    int w,t,tem_sum;
    while (m--) {
        cin >> w >> t;
        tem_sum = 0;
        int tem;
        if (w>t) {
            tem = t;t = w; w = tem;
        }
        tem_sum = sum[t - 1] - sum [w - 1];
        cout << min(tem_sum,sum[n]-tem_sum) << endl;
    }
    return 0;
}