PAT1046题解

Posted 2021-02-07 dcklm

#include<iostream>
#include<math.h>
using namespace std;
const int INF = 1e9;
const int MAXN = 3*1e5;
int cost[MAXN], val[MAXN];
int N, num, a, b; 
int main() {
    cin >> N;
    for(int i = 1; i <= N;i++) {
        cin >> num;
        val[i] = val[N+i] = num;
    }
    for(int i = 1; i <= 2*N; i++) {
        cost[i] = val[i] + cost[i-1];
    }
    cin >> num;
    while(num--) {
        cin >> a >> b;
        if(abs(a-b) == 1) {
            int temp = min(a,b);
            cout << val[temp] << endl;
        }else {
            if(a > b) swap(a,b);
            int temp = min(cost[b-1] - cost[a] + val[a], cost[N+a-1] - cost[b] + val[b]) ;
            cout << temp << endl;
        }
    }
}

对于循环数组，可以将数组复制一倍。 思路源自  首位相连的数组的最大连续子段和