2019湘潭校赛 E(答案区间维护)

题目传送
思路是始终维护西瓜数量的区间,即L代表目前可以达到的最少的,R是最多的,然后判断一下。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
int T, n, m, a[maxn];

int main() {
    for (scanf("%d", &T); T; T--) {
        int maxx = 0, lila = 0;
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            if (a[i] > maxx) {
                maxx = a[i];
                lila = i;
            }
        }
            
        int L = m, R = m, i = 0;
        while (233) {
            i = i % n + 1;
            if (i != lila) {
                if (R <= 0) {
                    puts("NO");
                    break;
                }
                L -= a[i], R--;
            } else {
                if (L <= 0) {
                    puts("YES");
                    break;
                }
                L -= a[i], R -= a[i];
            }
        }
    }
}
原文地址:https://www.cnblogs.com/AlphaWA/p/10897758.html