答案为n∗m−1+4n*m-1+4n∗m−1+4
443
试题B：灭鼠先锋
LLLV
试题C：求和
预计得分100%
思路：维护一个前缀sumsumsum即可。
总时间复杂度 O ( n ) O(n)O(n)
参考代码：
#include using namespace std;const int N = 2e5 + 5;int a[N];void solve(){int n;scanf("%d", &n);long long ans = 0, sum = 0;for (int i = 1; i 1;build(k * 2, l, mid);build(k * 2 + 1, mid + 1, r);pushup(k);}int query(int k, int l, int r){if (tr[k].l == l && tr[k].r == r)return tr[k].maxx;int mid = tr[k].l + tr[k].r >> 1;if (r mid)return query(k * 2 + 1, l, r);elsereturn max(query(k * 2, l, mid), query(k * 2 + 1, mid + 1, r));}void solve(){int n, m, x;scanf("%d %d %d", &n, &m, &x);for (int i = 1; i=2∗xsum>= 2*xsum>=2∗x即合法。总时间复杂度 O ( n l o g n ) O(nlogn)O(nlogn)
参考代码：
#include using namespace std;const int N = 3e5 + 5;int a[N];long long dp[N];// dp[i]表示最多能跳到位置i dp[i]次int n, x;bool check(int mid){long long sum = 0;for (int i = 1; i