PAT-1046 Shortest Distance

/**************************
//@Author: 3stone
//@ACM: PAT-A1046
//@Time: 18/1/19
//@IDE: VS2017
***************************/
#include<cstdio>
#include<iostream>
#include<algorithm>

#define maxSize 100005
using namespace std;

int main() {
    /*
    //N最大值可取10^5,M最大值可取10^4,因此若用嵌套循环的话,
    //最坏情况下可达10^9次操作,这在100ms内无法完成
    //【解决办法】每次输入,记录累加值
                //最短距离,无非就是对比一下 左走 还是 右走 嘛
    */
    int n, m, left, right, sum = 0;
    int dis[maxSize] = { 0 }, a[maxSize] = { 0 };
    //input distance
    scanf("%d", &n);
    if (n < 3 || n > maxSize) return 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        sum += a[i];
        dis[i] = sum;
    }
    //search
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &left, &right);
        if (left > right)//保证left < right
            swap(left, right);
        int temp = dis[right - 1] - dis[left - 1];
        printf("%d\n", min(temp, sum - temp));   
    }

    system("pause");
    return 0;
}

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转载自blog.csdn.net/qq_26398495/article/details/79109421