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

题意:给定n个点它们围成一个圆,给定相邻两个点的距离,给出m个询问,问任意两个点之间的最短距离

模拟即可.

#include <cstdio>#include <algorithm>using namespace std;int main() {    int n, m;    int sum[100010] = {0};    scanf("%d", &n);    for (int i = 1; i <= n; ++i) {        scanf("%d", sum + i);        sum[i] += sum[i - 1];    }    scanf("%d", &m);    while (m--) {        int a, b;        scanf("%d%d", &a, &b);        if (a > b) swap(a, b);        printf("%d\n", min(sum[b - 1] - sum[a - 1], sum[a - 1] + sum[n] - sum[b - 1]));    }    return 0;}