PAT甲级练习1046. Shortest Distance (20)

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 ... 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

计算出两个方向的距离，取小的那个

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <set>#include <string>#include <string.h>using namespace std;const int MAX=1e5+10;int sum[MAX];int getd(int a, int b, int c){int d;if(sum[a]>sum[b]) d=min(sum[a]-sum[b], c-sum[a]+sum[b]);else d=min(sum[b]-sum[a], c-sum[b]+sum[a]);return d;}int main(){int n, m, x, y;scanf("%d", &n);sum[1] = 0;for(int i=2; i<=n+1; i++){scanf("%d", &x);sum[i] = sum[i-1] + x;}scanf("%d", &m);for(int i=0; i<m; i++){scanf("%d %d", &x, &y);printf("%d\n",getd(x, y, sum[n+1]));}cin>>n;return 0;}