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  • PAT 甲级 1046 Shortest Distance

    https://pintia.cn/problem-sets/994805342720868352/problems/994805435700199424

    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, 10^5^]), followed by N integer distances D~1~ D~2~ ... 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 (<=10^4^), 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 10^7^.

    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

    代码:
    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    int a[maxn], dis[maxn];
    
    int main() {
        int N, sum = 0;
        scanf("%d", &N);
        for(int i = 1; i <= N; i ++) {
            scanf("%d", &a[i]);
            sum += a[i];
            dis[i] = sum;
        }
    
        int T, left, right, temp;
        scanf("%d", &T);
        for(int i = 1; i <= T; i ++) {
            scanf("%d%d", &left, &right);
            if(left > right) {
                swap(left, right);
            }
            temp = dis[right - 1] - dis[left - 1];
            printf("%d
    ", min(temp, sum - temp));
        }
    
        return 0;
     }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9439663.html
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