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
思路
- 如果设起点、终点分别为
x,y，顺时针的距离就是dis(x,y)，逆时针的距离就是dis(y,x)，比较就好了。如果是对每一组测试点都手动模拟的话会TLE - 所以需要优化，这题的本质是区间和的比较，我们可以另开一个数组记录区间和，那么每次查询的时候只要直接相减就好了
代码
#include<bits/stdc++.h>
using namespace std;
int a[100010] = {0};
int length[100010] = {0};
int main()
{
int n;
scanf("%d", &n);
int sum = 0;
for(int i=1;i<=n;i++)
{
scanf("%d", &a[i]);
sum += a[i];
length[i] = sum;
}
int m;
scanf("%d", &m);
int l, r;
int t; //暂时存储距离
while(m--)
{
scanf("%d %d", &l, &r);
if(l > r) swap(l, r);
t = length[r-1] - length[l-1];
printf("%d
", min(t, sum - t));
}
return 0;
}
引用
https://pintia.cn/problem-sets/994805342720868352/problems/994805435700199424